MBARARA UNIVERSITY OF SCIENCE AND TECHNOLOGY

P. O. BOX 1410. MBARARA

LECTURE NOTES: MTH1102 CALCULUS I

By

John Emenyu (PhD)

BSc./Ed (MUST, UGANDA), MA/MATHEMATICS (KU, USA), PhD (MUST)

 Outline

Sequences

The  − N () definition of a limit of a sequence. Theorems on limits of sequences and hence
techniques of evaluation of limits of sequences. series defined as partial sum of a sequence.
Ratio and comparision test for convergence.

Functions

Polynomials (Linear and Quadratic), rational, composite functions, transcendental functions,

logarithmic and exponential functions and their inverses.

Limits

Informal definition of limits functions and continuity, one sided limits, removable discontinuity.
Techniques and/or theorems for evaluation of limits. The formal  − δ definition of a limit:
applications to the definitions and properties of continuous functions. Use of the definition in
proofs and problems on limits and continuity.

Differentiation

Definition of a dirivative, continuity and differentiability. Rules and/or theorems for determin-
ing derivatives. Inverse functions: Their derivatives and graphs.

Differentials

Application to approximation, roles Theorem, mean value theorem (cauchy’s M.V.T), L’Hopital’s
rule, Taylor’s theorem and application.

References

The following are some of the recommended text books for use in this course.

1) Edwards and penny, Calculus and Analytic Geometry.

2) S. C. Agrawal and D. Pandry, Fundamentals of Differential Calculus.

3) K. G. Binmore, Mathematical Analysis.
4) Any other text on Calculus.

1
Chapter 1

Sequences

1.1 The Real line

We shall take real numbers as points along the real line which extends indefinitely in both
directions.
subset of R Notation Elements
Natural numbers N 1,2,3,4,. . .
Integers Z . . . , -2,-1,0,1,2,. . .
Rational numbers Q -2,-1,-1/2,0,
√ √ 1/2,1,. . .
Irrational numbers Q0 . . . , 2, 3, . . .

Definition 1.1.1. A sequence of real numbers is the function whose domain is a set of N. That
is, a sequence is a function f such that f : N → R.
Examples 1.1.2. The following are examples of sequences.

1) The set of negative integers form a sequence. Here the function is defined by f (x) = −x.
2) The set {1, 21 , 13 , 14 , . . . , } is a sequence with the function f (x) = x1 .
Exercise 1.1.3. Give four examples of sequences.
Notation 1.1.4. Note that a sequence is a list of real numbers denoted by {an } = {a1 , a2 , . . .}
where n ∈ N. Here f (n) = an for n ∈ N and an ∈ R.

1.1.1 The least upper bound (l.u.b)

An upper bound for a set S ⊂ R is a number a ∈ R such that x 6 a for all x ∈ S. If a set S has
an upper bound, then we say that it is bounded from above. A real number y is a least upper
bound of a set S ⊆ R if

(i) y is an upper bound for S.

(ii) a is any other upper bound for S then y ≤ a.
Example 1.1.5. Let S = {1, 2, 3}. The upper bound for S can be 3, 4, 5, etc. The least upper
bound (l.u.b) = 3. The l.u.b is commonly known as the supremum and abreviated as sup(for
example: sup{1, 2, 3} = 3).

2
1.1.2 The greatest lower bound (g.l.b)

A lower bound for a set S ⊂ R is a number b ∈ R such that b 6 s for all s ∈ S. If a set S has a
lower bound, we say it is bounded from below. A number x ∈ R is a g.l.b of the set S if

i) x is a lower bound for S.

ii) b is any other lower bound for S then b 6 x.

The g.l.b is known as infimum, abreviated as inf.

Remark 1.1.6. Let S be a subset of real numbers R.

i) The maximum of a set S is the largest (in value) element of S.

ii) The minimum of a set S if the smallest (in value) element of S.
iii) The l.u.b of a set S ⊂ R, if it has, is not necessarily in S. For example,

S = {x ∈ R : x < 0}

has l.u.b sup S = 0 but 0 ∈

/ S.
iv) A set S may not have a maximum (similarly, minimum) but if it has then the maximum
(minimum) is equal to suprmeum (infimum).

1.1.3 The supremum and infimum properties of subsets of Real num-

bers

Let S ⊂ R be non-empty. Then

i) S ⊂ R always has a l.u.b (or g.l.b) if it is finite.

ii) S ⊂ R has a l.u.b (or g.l.b) if it is bounded from above (or below).
Proposition 1.1.7. For any real x there is an integer n such that n > x.

Proof. We shall prove by way of contradiction. Assume we have a number x ∈ R for which the
assertion is wrong. Then n 6 x for each n ∈ Z so that the set of all integers Z is bounded from
above. Since the set of integers is non empty it has l.u.b, say a. But for any integer n, n + 1 is
also an integer. So (n + 1) < a ⇒ n < a − 1. Therefore a − 1 is also an upper bound for Z.
Since a − 1 < a, it means that a is not the least upper bound. Therefore, the original statement
must be true.

The following is the  − N () definition of a limit of a sequence.

Definition 1.1.8. A sequence {an } is said to converge to a real number L if for every  > 0
exists an integer N ∈ N such that for all n ≥ N , |an − L| < . We write lim an = L and call
n→∞
L the limit of the sequence {an }.
Remark 1.1.9. The Definition 1.1.8 says that an → L as n → ∞ by chosing n sufficiently
large.
Examples 1.1.10. Show that
 
1
i) lim 2+ = 2.
n→∞ n

3
 1
ii) lim = 0.
n→∞ n

Solution: Let  > 0 be given. Then:

i) We must find an integer N such that for every n ≥ N , |an − 2| < . But, |an − 2| =
|2 + n1 − 2| <  ⇒ | n1 | <  ⇒ n1 < . Therefore n > 1 . Thus, 1
 pickany fixed integer N >  .
1
Then for every n > N > 1 , |2 + n1 − 2| < . Hence lim 2 + = 2.
n→∞ n
ii) It suffices to find N ∈ N such that for all n > N |an − 0| < . But, |an − 0| = | n1 − 0| <
 ⇒ | n1 | <  ⇒ n1 < . Therefore, n > 1 . So, we can pick an N > 1 such that for every
1
n ≥ N > 1 , | n1 − 0| < . Hence lim = 0.
n→∞ n

Theorem 1.1.11. Let {an } and {bn } be convergent sequences of real numbers with limits a
and b respectively. Then

a) lim (an ± bn ) = a ± b.
n→∞
b) lim (an bn ) = ab.
n→∞
 
an a
c) lim = for b 6= 0.
n→∞ bn b
d) lim can = ca for c ∈ R.
n→∞

Proof. a) Given  > 0, we also have 2 > 0; so, we can find N1 ∈ N, such that |an − a| < 2 for
all n > N1 . Similarly, we can find an integer N2 ∈ N such that |bn − b| < 2 for all n ≥ N2 .
Let N = max{N1 , N2 }. Then, whenever n > N , have
|(an + bn ) − (a + b)| = |(an − a) + (bn − b)|
≤ |an − a| + |bn − b|
 
< + = .
2 2
Therefore given any  > 0, there exists N ∈ N such that |(an + bn ) − (a + b)| <  for all
n > N . Hence, lim (an + bn ) = a + b. For subtraction, it can be proved in a similar way.
n→∞
b) First observe, the triangular inequality gives |an bn − ab| = |an bn − an b + an b − ab| ≤
|an ||bn − b| + |b||an − a|. We also note that all convergent sequences are bounded and so
there exists a real number M > 0 such that |an | 6 M for all n. Note that our goal is to
determine some N ∈ N such that |an bn − ab| 6 |an ||bn − b| + |b||an − a| <  for all n > N .
To do this, let  > 0 be given. Then:

1) By convergence of {an } to a, choose N1 ∈ N such that |an − a| < 2|b|+1 for all n > N1 .

2) Similarly, by convergence of {bn } to b choose N2 ∈ N such that |bn − b| < 2M for all
n > N2 .
Let N = max{N1 , N2 }; it then follows from triangular inequality that |an bn −ab| 6 |an ||bn −
M |b|
b| + |b||an − a| < 2M + 2|b|+1 < 2 + 2 =  for all n > N . That is, lim (an bn ) = ab.
n→∞
1 1
c) It suffices to prove that lim = for b 6= 0. Now, let  > 0 be given. Choose N1 ∈ N
n→∞ bn b
such that |bn − b| < 2 |b|2 for all n > N1 . Then notice |bn | = |b − (b − bn )| > |b| − |b − bn |.
So, choosing N2 ∈ N such that |bn − b| < |b| 2 for all n > N2 makes |bn | > |b| − |b − bn | >
|b| − |b|
2 = |b|
2 . This implies that 1
|bn | ≤ 1
|b| . So, let N = max{N1 , N2 }. Then, for all
|b|2
n > N we have | b1n − 1b | = |b−b n| |b−bn |
|bbn = |b||bn | < |b| |b| = . So, it follows from part (b) that
2

    2
an 1 a
lim = lim an lim = for b 6= 0.
n→∞ bn n→∞ n→∞ bn b

4
 d) Let bn = c for all n ∈ N. Then, lim bn = c and since |bn − c| = 0 for all n ∈ N.
n→∞
Thus, for a given any  > 0, trivially we have |bn − c| < . Then apply part (b) to get
lim (bn an ) = lim (can ) = ca.
n→∞ n→∞

Theorem 1.1.12. (Sandwich or squeezing theorem). Suppose that lim an = L = lim cn . If

n→∞ n→∞
an 6 bn 6 cn for all n then lim bn = L.
n→∞

Proof. First note that |b − L| <  if and only if L −  < b < L + . Let  > 0 be given; then
choose N1 ∈ N such that for all n > N1 , |an − L| < . Similarly, choose N2 ∈ N such that for
all n > N2 , |cn − L| < . Let N = max{N1 , N2 }. Then, for n > N , we have that |an − L| < 
if and only if L −  < an < L + , and |cn − L| <  if and only if L −  < cn < L + ; so, this
implies that L −  < an 6 bn 6 cn < L + ; thus, L −  < bn < L +  for all n ∈ N. That is, for
all n > N we have |bn − L| < ; in other words, lim bn = L.
n→∞

(−1)n
Example 1.1.13. Show that lim cos n = 0.
n→∞ n2

−1 (−1)n 1 1
Solution: First notice that n2 6 n2 cos n 6 n2 and lim ± = 0. So, by Sandwich
n→∞ n2
n
(−1)
theorem we have that lim cos n = 0.
n→∞ n2
2n
Example 1.1.14. Find the limit of the sequence an = 5n−1 .

2n 2
Solution: Write an = 5n−1 = 1
5− n
. Then it is clear that

2  −1 2
−1
lim an = lim = 2 lim (5 − n ) =
n→∞ n→∞ 5 − n−1 n→∞ 5
.
Theorem 1.1.15. A sequence converges to a unique limit.

Proof. Suppose that lim an = L1 and lim an = L2 where L1 6= L2 and let  > 0 be given.
n→∞ n→∞
There exist


1) N1 ∈ N such that |an − L1 | < 2 whenever n > N1 .

2) N2 ∈ N such that |an − L2 | < 2 whenever n > N2 .

Let N = max{N1 , N2 }. Then for all n > N we have

 
(1.1.1) |L1 − L2 | = |an − L1 + an − L2 | 6 |an − L1 | + |an − L2 | < + = .
2 2
So, L1 = L2 since if L1 − L2 = ρ > 0, then the Eq. (1.1.1) would be false for some  > 0.
Exercise 1.1.16. Work out the following exercises.

1) Suppose that lim an = A. Prove that lim |an | = |A|.

n→∞ n→∞
2) Prove that {(−1)n an } diverges if lim an = A 6= 0.
n→∞
3) Suppose that lim an = a. Show that lim (an + c) = a + c for c ∈ R.
n→∞ n→∞

5
 4) Let {an } and {bn } be bounded sequences of real numbers. Show that lim sup(an + bn ) 6
n→∞
lim sup(an ) + lim sup(bn ).
n→∞ n→∞

Definition 1.1.17. (Monotone Sequences) A sequence {an } is said to be monotonically de-

creasing if an+1 6 an for all n = 1, 2, . . . and monotonically increasing if an+1 > an for all
n = 1, 2, . . ..
Examples 1.1.18. The sequence

1) { n1 } = {1, 12 , 13 , . . .} is monotonically decreasing.

2) {n2 } = {1, 4, 9, . . .} is monotonically increasing.
3) {(−1)n } = {−1, 1, −1, 1, . . .} is neither monotonically increasing nor monotonically de-
creasing.
Definition 1.1.19. A sequence {an } is said to be a Cauchy sequence if given  > 0, there
exists N ∈ N such that for all m, n > N , |an − am | < .
Proposition 1.1.20. Every convergent sequence is Cauchy.

Proof. Let an → L as n → ∞. Then for any given  > o, there exists N ∈ N such that
|an − L| < 2 whenever n > N . So for m, n > N we have |an − am | = |an − L + L − am | 6
|an − L| + |am − L| < 2 + 2 = . Hence for m, n > N , |an − am | < . Thus, any convergent
sequence is cauchy.
Proposition 1.1.21. Every cauchy sequence is bounded.
Exercise 1.1.22. Does every cauchy sequence converge? If not, give an example of a divergent
Cauchy Sequence.

1.2 Series

Definition 1.2.1. An infinte series is an expression of the form

∞
X
(1.2.1) an = a1 + a2 + ... + an + . . . ,
n=1

where an is an infinite sequence of real numbers. We call the number an the nth term of the
Series in Equation (1.2.1).
Example 1.2.2. The following are common examples of series.

x2 x4
P∞ (−1)n 2n
i) cos x = 1 − 2! + 4! − ... = n=0 (2n)! x .
x2 x3 xn
P∞
ii) ex = 1 + x + 2! + 3! + ... = n=0 n! .

Definition
P∞ 1.2.3. The sum Sn = a1 + a2 + a3 + ... + an of the first n terms of the series
k=0 a k is called the nth partial sum of the series.
Remark 1.2.4. It follows from the Definition 1.2.3 that each infinite series is associated with
an infinite sequence of partial sums {Sn }.

6
Definition 1.2.5. We say that the infinite series (1.2.1) converges (or is convergent) with sum
S provided that the limit of its sequence of partial sums, S = lim Sn exists and is finite; We
n→∞
∞
X
write S = an = lim Sn provided that this limit exists. Otherwise, we say that the series
n→∞
n=0
(1.2.1) diverges (or is divergent). If the series (1.2.1) diverges, then either it has no unique
limit or the limit of partial sums is infinite.
P∞
n
Example 1.2.6. Show that the series n=1 21 converges and find its sum.
P∞ 1 n 1 1 1
Solution: Consider n=1 ( 2 ) = 2 + 4 + 8 + . . .. Now take the first three partial sums
1 21 − 1 1 1 3 22 − 1 1 1 1 7 23 − 1
S1 = = , S2 = + = = , S3 = + + = = ,....
2 2 2 4 4 22 2 4 8 8 23
n
In general, we see that the nth partial sum is given by Sn = 2 2−1
n . Thus, the sum of the given

infinite series is given by

2n − 1
 
1
S = lim Sn ⇒ S = lim = lim 1 − = 1.
n→∞ n→∞ 2n n→∞ 2n
∞  n
X 1
That is, the series converges to 1.
n=1
2
∞
X
Example 1.2.7. Show that the series (−1)n+1 diverges.
n=1

P∞
Solution: Notice for the given series n=1 (−1)n+1 = 1 − 1 + 1 − 1 + 1 − 1 + . . ., the sequence
of partial sums {Sn } = {1, 0, 1, 0, 1, . . .}, that is Sn = 1 for odd values of n ∈ N and Sn = 0 for
even values of n ∈ N. This means that

1 if n is odd
S = lim Sn =
n→∞ 0 if n is even
so that the given series has no limit and therefore diverges.
∞
X 1 1 1 1
Example 1.2.8. Show that the infinite series 2−1
= + + + . . . converges and
n=1
4n 3 15 35
find its sum.

Solution: To show that the above series converges, we need a formula for the nth partial sum,
Sn so that we can evaluate the lim Sn . Now,
n→∞
 
1 1 1 1 1
an = 2 = = − .
4n − 1 (2n + 1)(2n − 1) 2 2n − 1 2n + 1
Then by partial fraction decomposition technique we have
         
1 1 1 1 1 1 1 1 1 1 1 1 1 1
Sn = 1− + − + − + − + ... + −
2 3 2 3 5 2 5 7 2 7 9 2 n−2 n
       
1 1 1 1 1 1 1 1 1 1 1
= 1− 6 + 6 −6 + 6 −6 + ... + 6 −
2 3 2 3 5 2 5 7 2 2n − 1 2n + 1
 
1 1 n
= 1− = .
2 2n + 1 2n + 1

∞
n 1 X 1 1
Thus, S = lim = . Hence the series is convergent, and 2−1
= .
n→∞ 2n + 1 2 4n 2
n=1

7
Remark 1.2.9. The nth partial sum Sn in this example is called a telescoping sum. In such
a sum, most terms cancel out except a few. This type of sum provides us with a way of finding
sums of certain types of series.

The series in examples (a) and (b) are examples of a much more common sort of series called
the geometric series.
Theorem 1.2.10. A necessary condition for the series (1.2.1) to converge is lim an = 0.
n→∞

P
Proof. Suppose that an converges to a limit S, i.e lim Sn = S where S ∈ R. Then
n→∞
lim Sn = S ⇒ lim Sn−1 = S. But an = Sn − Sn−1 . Therefore,
n→∞ n→∞

lim an = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = S − S = 0 = 0.

n→∞ n→∞ n→∞ n→∞
P
Thus if an converges, then lim an = 0.
n→∞

Remark 1.2.11. It is significant to note that Theorem (1.2.10) gives a necessary but not a
sufficient condition for an infinite series in Eq. (1.2.1) to converge.
P∞
Corollary 1.2.12. If lim an 6= 0, then the infinite series n=1 an diverges.
n→∞
∞
X
Example 1.2.13. The series arn−1 = a + ar + ar2 + . . . where a 6= 0, r 6= 0 is called a
n=1
geometric series. Show that the G.P converges if |r| < 1 and diverges if |r| ≥ 1.
Solution: Set Sn = a + ar + ar2 + . . . + arn−1 and Sn = (a − arn )/r − 1. Now if |r| < 1, then
a
rn → 0 as n → ∞. Thus, lim Sn = . Hence the geometric series converges if |r| < 1.
n→∞ 1−r
If |r| > 1, then rn → ∞ as n → ∞. By Cororally (1.2.12) the geometric series diverges.

If r = 1, then Sn = na and Sn → ±∞ as n → ∞.

If r = −1, the series a − a + a + . . . + (−1)n+1 a + . . . and a 6= 0 has no unique sum as n → ∞.

Therefore the geometric series converges if |r| < 1 and diverges if |r| ≥ 1.
∞
X 1 1 1
Example 1.2.14. Consider the harmonic series = 1 + + . . . + + . . .. By integral
n=1
n 2 n
test to be looked at, the harmonic series diverges though an = n1 → 0 as n → ∞. This justifies
the remark that the condition an → 0 for convergence of series in Eq. (1.2.1) is a necessary
condition but not a sufficient one.
∞
X ∞
X
Theorem 1.2.15. If c is a constant and series an converges, then so does can , and its
n=1 n=1
∞ ∞
X X P∞
limit is c an . If can diverges then so is n=1 can provided c 6= 0.
n=1 n=1
∞ ∞
X X P∞
Theorem 1.2.16. If an and bn are convergent infinite series then n=1 (an + bn ) is
n=1 n=1
∞
X ∞
X ∞
X
convergent and (an + bn ) = an + bn .
n=1 n=1 n=1

Proof. Suppose Sn = a1 + a2 + . . . + an , P tn = b 1 +
Pb2 + . . . + bn and wn = Sn + tn =
(a1 + b1 ) + (a2 + b2 ) + ... + (an + bn ). Since an and bn are convergent series, lim Sn and
n→∞
lim tn exist. Therefore lim wn = lim (Sn + tn ) = lim Sn + lim tn = S + t.
n→∞ n→∞ n→∞ n→∞ n→∞

8
1.2.1 Test for Convergence of Series of Positive terms

In this section we shall consider series whose terms are all postive. We note that if all terms of
the series are postive the the sequence of partial sums is non decreasing. Therefore, a necessary
and sufficient condition of an infinite series of postive terms to converge is that its sequence of
partial sums must have an upper bound.
X∞ X∞ X∞
Definition 1.2.17. Suppose that an and bn are two series. The series bn is said to
Pn n=1 n=1 n=1
dominate the series n=1 if bn > an for all postive integers n.
∞
X ∞
X
Theorem 1.2.18. (Comparision Test) Suppose that an and bn are series of postive
n=1 n=1
∞
X ∞
X
terms and that bn dominates bn so that for all n ∈ N, bn > an . If
n=1 n=1

∞
X ∞
X
i) bn is convergent, then an is also convergent.
n=1 n=1
X∞ ∞
X
ii) an is divergent, then an is also divergent.
n=1 n=1

∞
X n
Example 1.2.19. Show that the series n (n + 1)
converges.
n=1
2

n n
Solution: We note that since 0 < n+1 < 1 for all n ∈ N then 0 < 2n (n+1) < 21n . Hence, the
∞
X 1 ∞
X 1
P∞ n
n
dominates the series dis n=1 2n (n+1) . Now is a convergent geometric series
n=1
2 n=1
2n
∞
X n
with r = 21 and so it follows that the series n (n + 1)
is also convergent, by comparison
n=1
2
test.
∞
X 1 1 1 1
Example 1.2.20. The series p
= 1 + p + p + . . . + p + . . . where p is a constant is
n=1
n 2 3 n
called a p-series. By the integral test, the p-series

a) converges if p > 1 and

b) diverges if p ≤ 1.
∞
X 1
Example 1.2.21. Show that the series √ is divergent.
n=2
4n − 7

∞
X 1
Solution: Notice that √ 1 > 1
√ for all n ∈ N\{1}. Thus the series √ dominates
4n−7 2 n
n=2
4n − 7
∞ ∞
X 1 X 1 1
the series √ . But √ is a divergent p-series since p = 2 < 1. Hence the series
n=2
2 n n=2
n
∞
X 1
√ is also divergent (by comparision test).
n=2
4n − 7

9
 ∞
X 1
Example 1.2.22. Test the series for convergence.
n=0
n!

Solution: Observe that for all n > 1, n! = n(n − 1)(n − 2) . . . 3.2.1 > 2.2.2 . . . 2.2.1. That is,
∞
1 1
P∞ 1
P∞ X 1
for all n > 1, n! > 2n−1 ⇒ n! > 2n−1 . But, n=1 2n−1 = n=0 21n . Hence, the series
n=0
n!
∞ ∞
X 1 X 1
is dominated by the series n
, a convergent geometric series. Hence the series is
n=0
2 n=0
n!
also convergent, by comparison test.
P
Remark 1.2.23. If an in the series an whose convergence is to be determined is a fraction,
we discard all except the largest magnitude term in both its numerator and denominator. That
is, we let
largest magnitude term in the numerator
bn = .
largest magnitude term in the denominator
∞
X ∞
X
Theorem 1.2.24. (Limit Comparision Test) If an and bn are series of postive terms
n=1 n=1
an
and if lim = L > 0, then either both series converge or both diverge.
n→∞ bn

an
Proof. Since lim = L > 0, there is a postive integer N such that | abnn − L| < L
2 for all
n→∞ bn
n > N . That is,
L an L L an 3L
− < −L< ⇒ < < .
2 bn 2 2 bn 2
L 2
So for all n > N we have that 2 bn < an and 3L an < bn . Thus, the series

∞
X XL
i) an dominates the series bn .
n=1
2
∞ ∞
X X 2
ii) bn dominates the series an .
n=1 n=1
3L

∞
X ∞
X
It now follows from the Theorem (1.2.24) that either both the series an and bn converge
n=1 n=1
or both diverge.
∞
X 1
Example 1.2.25. Show that the series √ is divergent.
n=2
4n2 − 7

Solution: Let bn = √ 1 and an = 1

4n2 −7 n. Then

1
r
an n 4n2 − 7
lim = lim = lim = 2.
n→∞ bn n→∞ √ 1 n→∞ n2
4n2 −7

1
P
But, since the harmonic series n diverges, then it follows that the given series diverges also.
∞
X 3n2 + n
Exercise 1.2.26. Show that the series √ converges.
n=1
n4 + n

10
 ∞ ∞
X X an
Theorem 1.2.27. If an and bn are series of postive terms and if lim = 0 and
n=1 n=1
n→∞ bn
∞
X ∞
X
bn converges then an also converges.
n=1 n=1

Proof. For abnn → 0 as n → ∞, there exists a postive integer N such that for all n > N ,
| abnn − 0| <  or |an | < |bn |. But for all n > N we have that an > 0, bn > 0 and so 0 < an < bn
P∞ P∞ P∞
P∞all n ≥ N . Thus, n=1 bn dominates n=1 an and since n=1 bn converges, so also does
for
n=1 an .
∞
X
an
Remark 1.2.28. It is important to note that if the series bn diverges and bn → 0, then
n=1
1
Theorem (1.2.27) fails and no conclusion is possible. As an illustration, consider an = n2 ,
2
bn = n1 . Then abnn = 1/n 1
P P
1/n = n but an converges while bn diverges.
∞
X an+1
Theorem 1.2.29. (The Ratio Test) Let an be a series of postive terms and let lim =
n=1
n→∞ an
ρ exists. If

∞
X
a) ρ < 1, the series an converges,
n=1
X∞
b) ρ > 1, the series an diverges,
n=1
c) ρ = 1, the test is inconclusive.
∞
X n20
Example 1.2.30. Determine whether the series is convergent.
n=1
2n

n20
Solution: Let an = 2n . So
20 20
(n + 1)20 /2n+1
 
an+1 1 1 1 1 1
= = 1+ ⇒ lim 1+ = .
an n20 /2n 2 n n→∞ 2 n 2
∞
1
X n20
So, since ρ = 2, the ratio test says the series converges.
n−1
2n
∞
X n20
Exercise 1.2.31. Test the series for convergence.
n=0
n!

1.2.2 Alternating series and absolute convergence

Definition 1.2.32. (Alternating Series) An infinite series of the form

∞
X
(1.2.2) (−1)n+1 an = a1 − a2 + a3 − . . .
n=1

where an > 0 for n ∈ N is called an alternating series.

11
Theorem 1.2.33. (Alternating Series Test) The alternating series (1.2.2) converges if 0 <
an+1 < an for all n ∈ N, an → 0 as n → ∞.
∞
X 1
Example 1.2.34. Test the series (−1)n+1 for convergence.
n=1
n

Solution: Here an = n1 , an+1 = n+1 1 1

. So, 0 < n+1 < n1 . Also, n1 → 0 as n → ∞. Thus, the
P∞
given series satisfies condition of Theorem (1.2.33), hence the series n=1 (−1)n+1 n1 converges.
∞
X n
Exercise 1.2.35. Test the series (−1)n+1 for convergence.
n=1
2n − 1

Definition 1.2.36. The infinite series

∞
X
(1.2.3) an = a1 + a2 + a3 + . . .
n=1
P∞
is said to converge absolutely if |an | = |a1 | + |a2 | + |a3 | + . . . converges.
n=1
P∞
Example 1.2.37. Determine whether the series n=1 cos n
n2 converges.

Solution: Let an = cos n cos n 1

n2 , then |an | = | n2 | 6 n2 for all n > 1. Hence, the convergent series
P∞ 1 P∞ | cos n|
n=1 n2 dominatesP the postive term series n=1 n2 . So it follows from the comparision
∞
test that the series n=1 cos n
n2 converges. Thus, the given series is absolutely convergent and
therefore converges.
Definition 1.2.38. If the series is convergent but not absolutely convergent, then it is said to
be conditionally convergent.
P∞
As an illustration of conditionally
P∞ convergent series n=1 (−1)n+1 n1 does not converge abso-
lutely since the series n=1 |(−1)n+1 n1 | = 1 + 12 + 13 + . . . the harmonic series diverges.
Remark 1.2.39.
P It is important to note that absolute
P convergence implies convergence i.e if
the series |an | convergesPthen so doesP the series an but the converse does not hold true.
Besides, if the two series |an | and an converge, it does not necessarily mean they both
converge to the same limit.
Theorem 1.2.40. If a series (1.2.3) converges abosolutely then it converges.
∞
X 10 sin nπ
6
Exercise 1.2.41. Discuss the convergence of the series .
n=1
n
p
Theorem 1.2.42. (The nth Root Test) Suppose that the limit ρ = lim n
|an | exists. Then
n→∞
∞
X
the series an
n=1

a) converges absolutely if ρ < 1.

b) diverges if ρ > 1.
c) the test is inconclusive if ρ = 1.
P∞ 1 1 1
Example 1.2.43. Consider the series n=1 2n+(−1)n =1+ 8 + 4 + . . .. If n is even then,
n
an+1 2n+(−1) 2n+1
= n+1+(−1)n+1 = n = 2
an 2 2

12
and if n is odd, then
an+1 2n−1 1
= n+1 = .
an 2 8
That is (
an+1 2, if n is even
= 1
.
an 8, if n is odd
Thus, (
an+1 2, if n is even
ρ = lim | |= 1
.
n→∞ an 8, if n is odd
Therefore the ratio test fails at this stage since ρ does not exist. However by the root test
1
(1.2.4) ρ = lim |an |1/n = lim | |1/n
n→∞ n→∞ 2n+(−1)n
1 1 1 1
= lim n+(−1)n
= lim n . (−1)n = .
n→∞
2 n
n→∞ 2 n
2 n 2
Therefore the series converge absolutely by the root test.
Exercise 1.2.44. Test whether the following series converge or diverge.

10 n
P∞

a) n=1 n .
P∞ (−1)n+1
b) n=1 2n .

1.2.3 The Power Series

Definition 1.2.45. If {an } is a sequence of the series

∞
X
(1.2.5) an xn = a0 + a1 x + a2 x2 + ... + an xn + . . .
n=0

is called a power series in x.

Remark 1.2.46. The substitution of x into (1.2.5) with any particular real number yields an
infinite series of real numbers. The series therefore must converge if x = 0 and for some non
zerovalues of x, and diverges for other values of x.
Definition 1.2.47. The set of numbers for which the power series converges is called the
convergent set for the power series.
Remark 1.2.48. The ratio test (1.2.29) is effective in determining the convergent set of the
power series (1.2.5). Therefore, setting Un = an xn , we have
Un+1 an+1 xn+1 an+1
ρ0 = lim | | = lim | | = |x| lim | | = ρ|x|,
n→∞ Un n→∞ an xn n→∞ an
where ρ = limn→∞ | aan+1
n
| provided this limit exists. By the ratio test, the power series will

1
(a) converge if ρ|x| < 1 i.e |x| < ρ = r,
(b) diverge if |x| > r,
(c) be inconclusive if |x| = r.

Thus, any power series in x to which we can apply the ratio test will have for its convergence
set one of the following;

13
 (i) All points in an open interval (−r, r) whose centre is the origin and possibly one or both
end points. We call r the radius of convergence and the interval (−r, r), the interval of
convergence.
(ii) The origin alone in this case the series diverges for |x| > 0 and we take r = 0.
(iii) All real numbers. Here the interval of convergence is (−∞, ∞).
Example 1.2.49. Find the convergent set of the power series
∞
X
n!xn = 1 + x + 2x2 + 6x2 + . . . + n!xn + . . . .
n=0

Solution: Using the ratio test;

Un+1 (n + 1)!xn+1
(1.2.6) ρ0 = lim | | = lim | |
n→∞ Un n→∞ n!xn
(
0, if x = 0
= |x| lim (n + 1) =
n→∞ ∞, if x 6= 0.

Now

(i) for x = 0 ρ0 = 0 < 1, the sries converges.

(ii) for x 6= 0, ρ0 = ∞ and the series diverges.

Thus the given power series in x converges only for x = 0 and diverges for any other number.
Example 1.2.50. Show that the power series
∞
X xn x2
=1+x+ + ...
n=0
n! 2!

converges for all real numbers x.

Solution: Using the ratio test

Un+1 |x|
ρ0 = lim | | = lim = 0|x| = 0.
n→∞ Un n→∞ n + 1

Thus for all numbers x, ρ0 < 1 the series converges. Hence the given power series converges for
all real numbers x.
Exercise 1.2.51. Prove that if k is an arbitrary real number and −1 < x < 1 then
k(k − 1) . . . (k − n)xn
lim = 0.
n→∞ n!
P∞
Theorem 1.2.52. If the series n=0 an xn converges for a number x1 6= 0, then it convergs
absolutely for all numbers x such that |x| < |x1 |.
P∞
Corollary 1.2.53. If the power series n=0 an xn diverges for a number x2 then it diverges
for all numbers x such that |x| > |x2 |.
Remark 1.2.54. It follows from Theorem 1.2.52 and Corollary 1.2.53 that if a power series
converges at a point x 6= 0 then either the series converges absolutely everywhere or there is a
positive number r such that the series converges absolutely throughout the open interval (−r, r)
and diverges everywhere outside the closed interval [−r, r].

14
Remark 1.2.55. Theorem 1.2.52 and Corollary 1.2.53 say nothing about the convergence of the
power sries 1.2.5 at the end points of the interval of convergence. For this reason, we therefore
substitute x = ±r into the original series 1.2.5 to obtain an infinite series with constant terms
whose convergence (or divergence) can easily be determined by the tests discussed above.
Example 1.2.56. Find the radius, interval of convergence set and the convergence of the power
series
∞
X xn
.
n=0
n2n

xn xn+1
Solution: Applying the ratio test, with Un = n2n ⇒ Un+1 = (n+1)2n+1 we have that

xn+1 n2n |x| n |x|

ρ0 = lim | n+1
. n |= lim = .
n→∞ (n + 1)2 x 2 n→∞ n + 1 2

Thus ρ0 < 1 if |x| < 2. The radius of convergence is therefore r = 2. So the given series
converges when |x| < 2 and diverges when |x| > 2. Now if

(i) x = 2 we obtain the harmonic series

∞ ∞
X 1 n X1 1 1
n
2 = = 1 + + + ...
n=1
n2 n=1
n 2 3

which diverges.
(ii) x = −2, we obtain the alternating harmonic series
∞ ∞
X (−2)n X (−1)n 1 1
n
= = −1 + − + . . .
n=1
n2 n=1
n 2 3

which converges.

Therefore the convergece set of the above series is [−2, 2).

Example 1.2.57. Find the interval of convergence of the power series
∞
X
nn xn .
n=0

Solution: Setting Un = nn xn , then

Un+1 (n + 1)n+1 xn+1

(1.2.7) lim | | = lim | |
n→∞ Un n→∞ nn xn
1
= |x| lim n(1 + )n+1 = ∞|x| ∀x 6= 0.
n→∞ n
Thus the given series converges only at x = 0 and its interval of convergence consists of a single
point x = 0.
Example 1.2.58. Find the interval of convergence of the power series
∞
X 2n n
x .
n=0
n!

15
Solution:
Un+1 2n+1 xn+1 n!
(1.2.8) ρ0 = lim | | = lim | |
n→∞ Un n→∞ (n + 1)! 2n xn
2
= |x| lim = 0|x| ∀x.
n→∞ n + 1

1
Thus ρ, < 1 if |x| < 0 = ∞. Hence the interval of convergence is (−∞, ∞).
Exercise 1.2.59. Find the interval of convergence of the series

∞
X xn
(i) .
n=0
n+2
∞
X 2n xn
(ii) .
n=0
(n + 1)!

Definition 1.2.60. If {an } is a sequence, then the expression

∞
X
(1.2.9) an (x − b)n = a0 + a1 (x − b) + a2 (x − b)2 + . . .
n=0

is called a power series in (x − b), b ∈ R.

Remark 1.2.61. The power series (1.2.9) in x − b) can be obtained from a power series (1.2.5)
in x by the transltion x = x0 − b. All that applies to 1.5.0 equally applies to (1.2.9). The
interval of convergence is now (b − r, b + r) wit the centre at point b. If the power series in
(x − b) converges only for x = b we set r = 0 and if it converges for all real numbers then we
set r = ∞.
Example 1.2.62. Find the interval of convergence and the convergence set of the series in
x − 2;
∞
X (−1)n (x − 2)n x − 2 (x − 2)2
=1− + − ....
n=0
n+1 2 3

Solution: Using the ratio test;

Un+1
(1.2.10) ρ0 = lim | |
n→∞ Un
(−1)n+1 (x − 2)n+1 n+1
= lim | |
n→∞ (n + 2) (−1)n (x − 2)n
n+1
= |x − 2| lim = |x − 2|.
n→∞ n + 2

For convergence, ρ0 < 1 implies |x − 2| < 1 ⇒ −1 < x − 2 < 1 ⇒ 1 < x < 3. When x = 1 we
get the series
∞ ∞ ∞
X (−1)n (−1)n X 1 X 1
= = ,
n=0
n+1 n=0
n + 1 n=1 n
a divergent harmonic series.

When x = 3, we get
∞
X (−1)n
,
n=0
n+1
a convergent alternating series. Therefore, the convergence set of this power series in (x − 2)
is [1,3].

16
Example 1.2.63. Determine te convergence set of the power series
∞
X (−1)n (x − 2)n
√ .
n=0
4n n

Solution:
Un+1
(1.2.11) ρ0 = lim | |
n→∞ Un
√
(−1)n+1 (x − 2)n+1 4n n
= lim | √ |
n→∞ 4n+1 n + 1 (−1)n (x − 2)n
r
|x − 2| n |x − 2|
= lim = .
4 n→∞ n + 1 4

Therefore, the radius of convergence is r = 4 since convergence is possible only if |x − 2| < 4.

Now |x − 2| < 4 =⇒ −2 < x < 6. Hence the series converges when −2 < x < 6 and diverges
when x < −2 and x > 6.
∞
X 1
When x = −2 the series reduces to the divergent p-series √ and when x = 6 the series
n=1
n
∞
X (−1)n
reduces to the convergent alternating series √ . Thus, the convergent set of the power
n=1
n
series is [−2, 6].
Exercise 1.2.64. Find the convergence set of the following power series.

∞
x x2 x3 x4 X (−1)n xn
(1) − + − + ... = .
1.2 2.3 3.4 4.5 n=1
n(n + 1)
x2 x4 x6
(2) 1 − 2! + 4! − 6! + . . ..
∞
X
(3) (−1)n n2 xn .
n=1
X∞
(4) (5x − 3)n .
n=1
∞
X 2n (x − 3)n
(5) .
n=1
n2

17
Chapter 2

Functions

2.1 Introduction

Let A and B be non-empty sets; x and y be variable quantities in A and B respectively. Suppose
that a choice of value of x ∈ A completely determines the corresponding choice of value of y ∈ B.
Definition 2.1.1. A rule that assigns to each x ∈ A a definite value y ∈ B is called a function.
A set A is called the domain of definition of a function and set B is called the range of a function.
Notation 2.1.2. Functions are often denoted by letters such as f , g, e.t.c. The domain of a
function f is denoted by D(f ) and its range by R(f ). The value corresponding to x under the
function f is denoted by f (x).
Remark 2.1.3. (a) A function is often called a mapping and is said to map its domain to
its range. Thus if f : A → B is a function (a map) then R(f ) ⊂ B and D(f ) ⊂ A. In
general, it is possible that under the mapping f , B has some elements which are not the
images of D(f ) ⊂ A. Elements in A in this case are called objects.
(b) If x ∈ D(f ), then f (x) ∈ R(f ). f (x) reads “f of x” but not “f times x”.
Example 2.1.4. The formula ν = l3 is a rule or a correspondence (a function) between a set
of postive numbers l and a set of positive numbers ν. For example

Domain D(ν) Range R(ν)

1 1
2 8
3 27

where ν is the rule that assigns a value of l to a value of ν i.e f (l) = ν = l3 .

2.1.1 Special Functions

Definition 2.1.5. (Polynomial Functions) The function Pn (x) defined by

(2.1.1) Pn (x) = an xn + an−1 xn−1 + ... + a1 x + a0 ,

where an 6= 0, n ∈ N.

18
Remark 2.1.6. We make the following observations about polynomial functions.

(a) If an = 0 and n = 0, the degree of the polynomial function (2.1.1) is not defined. This
type of polynomial function without degree is called the zero function and its given by
P0 (x) = f (x) = 0.
(b) If in (2.1.1) a0 and all other coefficients are equal to zero then (2.1.1)) reduces to a function
P0 (x) = a0 called a constant function whose range consists of a single number i.e for all
x, P0 (x) = f (x) = a0 .
(c) If a1 = 1 and a0 = a1 = a2 = . . . an−1 = an = 0, then (2.1.1) yields a polynomial function
P1 (x) = x called an identity function.
(d) If a1 6= 0, then (2.1.1) gives P1 (x) = a1 x + a0 . This polynomial is called a linear function.
(e) Polynomials of degrees two, three etc are respectively called quadratic, cubic functions etc.
Remark 2.1.7. The domain of every polynomial functions (a) to (e) is a subset of R.
Definition 2.1.8. A rational function is a function which can be expressed as a quotient of
two polynomial functions. It is defined by

an xn + an−1 xn−1 + . . . + a1 x + a0 Pn (x)

(2.1.2) f (x) = m m−1
= ,
bm x + bm−1 x + . . . + b1 x + a0 Qm (x)

for all values of x for which the denominators Qm (x) 6= 0 i.e D(f ) = R \ {x : Qm (x) = 0}.
Definition 2.1.9. Afunction defined by y = f (x) i called an algebric function iff it satisfies
identically an equation of the form

(2.1.3) Pn (x)y n + Pn−1 (x)y n−1 + . . . + p1 (x)y + p0 (x) = 0,

where Pi (x), i = 0, 1, 2, . . . , n are polynomials and n is a postive integer.

Definition 2.1.10. A transcendental function is any function which is not algebric. Examples
of such functions are y = sin x, y = ex , y = loge x etc.
Definition 2.1.11. The greatest integer function denoted by [x] is defined to be the greatest
integer that is les or equal to x. Example: [π] = 3.
Definition 2.1.12. The function defined by

−x
 if x < 0,
f (x) = |x| = 0 if x = 0,

x if x > 0,


is called the absolute value function.

Definition 2.1.13. (The hyperbolic sine and hyperbolic cosine) These are defined by sinh x =
ex −e−x x −x P∞ n 2

2 and cosh x = e +e
2 , where ex = n=0 xn! = 1 + x + x2! + . . ..
Remark 2.1.14. From the above hyperbolic functions, other hyperbolic functions can be ob-
tained (tanh x, and coth x).

2.1.2 Operations on Functions

In this subsection, we define certain operations on functions and all the functions resulting from
such operations i.e the sum, product etc of the original functions.
Definition 2.1.15. Let f and g be function with domains D(f ) and D(g) respectively, then

19
 (i) (f + g)(x) = f (x) + g(x) and (f − g)(x) = f (x) − g(x).
(ii) (f.g)(x) = f (x).g(x).
f (x)
(iii) ( fg )(x) = g(x) , g(x) 6= 0.

In each of the above operations, the domain consists of all values of

 xcommon to both D(f )
and D(g) i.e D(f + g) = D(f − g) = D(f.g) = D(f ) ∩ D(g) and D fg = D(f ) ∩ D(g) − {x :
g(x) = 0}.
1 √
Example 2.1.16. Let f (x) = x−3 and g(x) = x, then their sum is

1 √
(f + g)(x) = f (x) + g(x) = + x,
x−3
and D(f ) = R|{3}, D(g) = [0, ∞). So, D(f + g) = R \ {3} ∩ [0, ∞) = [0, ∞)|{3}.
√
Example 2.1.17. Find the domain of the function h(x) = log101(1−x) + x + 2.

√
Solution: Let f (x) = log101(1−x) and g(x) = x + 2. Thus D(f ) = (−∞, 1]/{0} and D(g) =
[−2, ∞). So D(h) = D(f + g) = [(−∞, 1]/{0} ∩ [−2, ∞) = [−2, 1]/{0}.
x
Exercise 2.1.18. Let f (x) = x+5 and g(x) = −2x3 . Find

(i) (f.g)(x).
f
(ii) g (x).
(iii) (f − g)(x).

Also determine domains in each of the above cases.

Definition 2.1.19. Suppose f and g are two functions with domain D(f ) and D(g) respectively.
The function f ◦ g defined by (f ◦ g)(x) = f (g(x)) is called a compoite function and its domain
is defined by D(f ◦ g) = {x : x ∈ D(g) and g(x) ∈ D(f )}.
Remark 2.1.20. (a) It should be noted that the composite function f ◦ g is defined only for
those x ∈ D(g) for which g(x) ∈ D(f ). In otherwords R(g) ⊆ D(f ) and consequently
D(f ◦ g) ⊂ D(g).

(b) The operation of composition gives another and very important way of constructing
√ new
functions from old ones. For example, let f (x) = (x26x
−9) and g(x) = 3x, then the
√
2 3x
operation of composition gives a new function F = f ◦ g = x−3 . The domain of F is
[0, 3) ∪ (3, ∞) = [0, ∞) − {3}.
(c) The composition function f ◦ g and g ◦ f are both different from each other. There should
be no confusion between (f × g)(x) and (f ◦ g)(x).
√
Example 2.1.21. Find the composition function f ◦ g and g ◦ f for the functions f (x) = x
and g(x) = x2 + 1. Also find D(f ◦ g) and D(g ◦ f .

20
 √
Solution: (f ◦ g)(x) = f (g(x)) = x2 + 1 (g ◦ f )(x) = g(f (x)) = x + 1. We observe that
D(f ) = R+ and D(g) = R. D(f ◦ g) = {x : x ∈ D(g) and g(x) ∈ D(f )} = R but, D(g ◦ f =
{x : x ∈ D(f ) and f (x) ∈ D(g)} = R+ .
Remark 2.1.22. Example 2.1.21 illustrates the ideas in Remarks 2.1.20 ((a), (c)).

2.1.3 Equality of Functions

Two functions f and g are said to be equal if and only if for all x ∈ D(f ) and x ∈ D(g), then

(a) D(f ) = D(g).

(b) f (x) = g(x).
Remark 2.1.23. It is important to note here that two functions with the same rule defining
them but different domains of definitio are not equal. For example, functions defined by f (x) =
x + 3 and g(x) = (x2 − 3)/(x − 3) are unequal since D(f ) = R 6= D(g) = R \ {3. But, if D(f )
is modified to exclude 3 then f and g becomes equal.

2.1.4 The Inverse of a Function

Recall that a function f is a rule that assigns each element in a domain of f , D(f ), one and
only one value in R(f ). If f has an additional property that no two distinct elements in D(f )
have the same value in R(f ) then the function f i said to be one to one.
Definition 2.1.24. A function f with domain D(f ) is said to be one to one iff x1 , x2 ∈ D(f ),
x1 6= x2 implies f (x1 ) 6= f (x2 ).
Definition 2.1.25. If f is one to one function then the function g defined by x = g(y) iff
y = f (x); g is called the inverse of f . The domain of g is the range of f and the range of g is
the domain of f .
Diagramatical illustration

Remark 2.1.26. The inverse of function f is denoted by f −1 . Generally the function and its
inverse are naturally opposite to one another i.e if f −1 is the inverse of f , then f is also the
inverse of f −1 . Furthermore, each pair of inverse function f and f −1 has the property that
(f ◦ f −1 )(x) = x for x ∈ D(f −1 ) and (f −1 ◦ f )(x) = x for all x ∈ D(f ).
Theorem 2.1.27. A function f has an inverse iff f is one to one i.e f (x1 ) 6= f (x2 ) if x1 6= x2 .
If a function f defined by y = f (x) has an inverse g, then the function g is obtained by solving
y = f (x) for x = g(y).
1
Example 2.1.28. Find the inverse function of f defined by f (x) = x2 +1 , x ≥ 0.

21
Solution: Given y = f (x) = x21+1 . Therefore y = x21+1 Implies that y1 = x2 + 1. Thus
q
x = 1−y y = g(y). Note that we take a postive square root because x ≥ o. The domain D(g)
is the range of f i.e R(f ) = D(g) = (0, 1]. Hence taking x as the indepedent variable, we find
that g is the function defined by
r
1−x
g(x) = , x ∈ (0, 1].
x
Exercise 2.1.29. Find the inverse function of the following functions;

(i) f (x) = 2x − 5,
x3
(ii) f (x) = 2 + 1.
√
(iii) f (x) = x − 2.
(iv) f (x) = √1 .
x−3

In each case above, state the domain and range of f −1 and furthermore sketch the graphs of
y = f (x) and y = f −1 (x) on the same axis.

2.1.5 Even and Odd Functions

Definition 2.1.30. Let f be a function with its domain D(f ). If for every x ∈ D(f ), −x also
belongs to D(f ) then the function f is said to be:

(i) even if f (−x) = f (x).

(ii) odd if f (−x) = −f (x),

2.1.6 Geometric Concepts of Odd and Even Functions

The Definition (2.1.30) implies that the graph of an even function is symmetrical about the
y-axis while the graph of an odd function is symmetrical about the origin.
Graphs

Example 2.1.31. Test the following functions for evenness and oddness.

(i) f (x) = x2 , D(f ) = R.

(ii) g(x) = x + x1 , D(g) = R \ {0}.

22
Solution: We note that for every x ∈ D(f ), −x ∈ D(f ). Thus f (x) = x2 = (−x)2 = f (−x).
Hence f (x) = x2 is even. Similary, for every x ∈ D(g), −x ∈ D(g), then g(−x) = −x − x1 =
−(x + x1 ) = −g(x). Hence g(x) = x + x1 is odd.
Example 2.1.32. Prove that the product of an odd function and even function is odd.

Proof. Let f be an even function so that if every x ∈ D(f ), −x also belongs to D(f ) and
f (x) = f (−x). Similarly, let g be an odd function, i.e., for every x ∈ D(f ), −x ∈ D(g) and
g(−x) = −g(x). Now define the product of f and g as (f.g)(x) = f (x).g(x) where D(f.g) =
D(f ) ∩ D(g).

To discuss the oddness and evenness of f.g we first ensure that x ∈ D(f.g) ⇒ −x ∈ D(f.g).
Thus, let x ∈ D(f.g). Then x ∈ D(f ) ∩ D(g) ⇒ x ∈ D(f ) and x ∈ D(g). But since f is even
then −x ∈ D(f ). Similarly, because g is odd, −x ∈ D(g) if x ∈ D(g) ⇒ −x ∈ D(f.g).

Also, we have that

(f.g)(−x) = f (−x).g(−x) (Def. of product functions)

= f (x). − g(x) (Defs. of even and odd functions)
= −(f (x).g(x))
= −(f.g)(x).

Hence, the product of an even function and an odd function is odd.

Remark 2.1.33. (i) It must be carefully noted that there may exist functions which are
neither odd nor even, e.g f (x) = x(x − 1), D(f ) = R is neither even nor odd since
f (−x) = x(x + 1) 6= f (x) or − f (x).
(ii) There doesn’t, in general, exist any function which is both even and odd. However the
exceotion here is a zero function f (x) = 0 with f (x) = R. The graph of such a function if
x-axis

Exercise 2.1.34. (1) Find f ◦ g and g ◦ f and determine the domain in each of the following
functions.
(i) f (x) = x + 1, g(x) = 2x
( (
x ifx < 0, x if − 3x < 0,
(ii) f (x) = , g(x) = .
2x if x ≥ 0 5x if x ≥ 0
(2) If f1 : D1 → R and f2 : D2 → R are two functions, find the maximum domain of f1 ◦ f2 .
Find (f1 f2 )(x) if f1 (x) = |x|.
(3) Test each of the following functions for evenness and oddness
(i) f (x) = |x|.
(ii) f (x) = x(x + 1)(x + 2).
(4) Prove the following statements;

23
(a) the product of an odd and even function is odd.
(b) the sum of 2 even functions is an even function.
(c) the sum of 2 odd functions is an odd function.
(d) the product of two even functions is an even function.

24
Chapter 3

Limits and continuity of functions

3.1 Intuitional Approach

Consider the function: y = 2x + 5. It is a simple matter to see that when x = 3, y = f (3) = 11.
Suppose that for some reason, we are interested in how the functiona when x is closed to 3.
What do we mean by x close x to 3? Here, the values such as 2.98, 2.99, 2.999, etc which are
less than 3 are said to be closed to 3. On the other hand, values such as 3.02, 3.01, 3.001, etc
which are greater than 3 can also be considered to be closed to 3.

We shall have to consider the behaviour of the fucntion for all values of x which are closed to
3 from both sides, i.e., close from the left (values less than 3) and closed from the right (values
greater than 3). The Table A below gives the values of the function given by y = f (x) = 2x + 5
when x approaches 3 from the left and the Table B provides the information for the case when
x approaches 3 from the right.

Table A Table B
x y x y
2.5000 10.000 3.5000 12.0000
2.7000 10.4000 3.2000 11.4000
2.9000 10.8000 3.1000 11.2000
2.9900 10.9800 3.0100 11.0200
2.9990 10.9980 3.0010 11.0020
2.9990 10.9990 3.0001 11.0002

Table A tells us that the value of the function y = f (x) = 2x + 5 reaches closely to 11 as x
approaches 3 from the left. This statement is abbreviated to f (x) → 11 as x → 3− and is read
“f(x) tends to 11 as x tends to 3 from the left”. In a similar way the information provided by
the Table B is written as f (x) → 11 as x → 3+ and is read “f(x) tends to 11 as x tends to 3
from the right”.
Notation 3.1.1. In the limit notation, we shall write

(i) f (x) → 11 as x → 3− as lim− f (x) = 11.

x→3
+
(ii) f (x) → 11 as x → 3 as lim+ f (x) = 11.
x→3

25
Remark 3.1.2. Since f (x) → 11 irrespective of the from which side x approaches 3, we can in
such a case drop the positive and negative signs and write the whole information as f (x) → 11
as x → 3. We then say the limit of f (x) as x → 3 is 11 and write lim f (x) = 11. This means
x→3
that 2x + 5 approaches 11 as x approaches 3.

3.1.1 Example of a limit of a rational function

Consider a real valued function f (x) = (x2 −9)/(x−3). This function is not defined when x = 3,
since for this value, the denominator reduces to zero and division by zero is not permitted. We
can therefore use the idea of limits to study the behaviour of the function for points close to
x = 3 from either sides. Since by elementary algebra,
x2 − 9 (x − 3)(x + 3)
= = x + 3 for x 6= 3.
x−3 x−3
Thus, we have that
x2 − 9
lim = lim (x + 3) = 3 + 3 = 6.
x→3 x − 3 x→3

Graph

Definition 3.1.3. If a real valued function f (x) gets arbitrarily close to some number ‘L’ as
x gets close to some number ‘a’ (from either directions) without ever being equal a then we say
that the limit of f (x) is ‘L’ as x tends to ‘a’ and write lim f (x) = L.
x→a

3.2 A Mathematical Approach

Consider again the function f (x) = (x2 − 9)/(x − 3) for x 6= 3, that is,
(x − 3)(x + 3)
(3.2.1) f (x) = = x + 3.
x−3
From the graph above, we observe by geometrical considerations that we can make f (x) ap-
proaches 6 as closely as we wish by making x close to 3, but never equals 3. Suppose we want
to make f (x) as close as within 0.01 from 6, that is to say, 5.99 < f (x) < 6.01 of

(3.2.2) |f (x) − 6| < 0.01.

Now, the question is, how close should we take x close to 3 to make the Inequality 3.2.2 true?
From Equation 3.2.1,

(3.2.3) |f (x) − 6| = |x − 3|.

Thus, if 0 < |x − 3| < 0.01 then from Equation 3.2.3, |f (x) − 6| < 0.01.

Conclusion: Choosing some number 0.01, we can get a number 0.01 such that for every x
that is 0 < |x − 3| < 0.01, we get |f (x) − 6| < 0.01.

26
Remark 3.2.1. We make the following observations.

(1) There is nothing special about the choice of the number 0.01 in the example above. We
could have chosen 0.05, 0.001 or any other number  (epsilon) that is greater than zero and
by the same process as in the example above we could have obtained a number δ (delta)
bigger than zero depending upon  such that |f (x) − 6| <  whenever 0 < |x − 3| < δ.
(2) In the example above we started with the choice of  = 0.01 and arrived at δ = 0.01 so that
x2 − 9
− 6 <  = 0.01
x−3
whenever 0 < |x − 3| <= 0.01. It must be carefully noted that the value of δ is not always
equal to the value of  as is the case in this example. In general, it may not be equal to
2
−18
. For example, consider the function, f (x) = 2xx−3 for x 6= 3. Then for any arbitrary
choice of  > 0, we have δ = /2 such that
2x2 − 18
− 12 < 
x−3
whenever 0 < |x − 3| < δ = /2. In particular if  = 0.01 then δ = 0.005. Generally,
2x2 − 18 2(x2 − 9) (x − 3)(x + 3)
(3.2.4) − 12 = =2 −6
x−3 x−3 x−3
= 2|x + 3 − 6| = 2|x − 3|.
Therefore,
2x2 − 18
− 12 <  ⇒ 2|x − 3| < 
x−3
So that 0 < |x − 3| < /2 = δ.
Definition 3.2.2. Let f (x) be a real valued function defined on an open interval  containing  a
real number ‘a’. We say that the number ‘L’ is the limit of f (x) as x tends to ‘a’ lim f (x) = L
x→a
provided that given any  > 0 there exists a number δ > 0 (depending on ) such that |f (x)−L| <
 whenever 0 < |x − a| < δ.
Remark 3.2.3. In general, the choice of δ depends upon the choice of . We do not require
that there should existsome δ which works for all  but rather we wish that for each  > 0 there
exists a δ which works for it.

Recall that the identity |x − a| < δ means that −δ < x − a and x − a < δ. These two inequalities
put together are equivalent to a − δ < x < a + δ. The open interval a − δ, a + δ) with a as its
middle point is called the δ-neighbourhood of a and is denoted by Nδ (a).

If the point ‘a’ itself is excluded from the interval, it is called a deleted neighbourhood of ‘a’ and
it is denoted by Nδ∗ (a).

In terms of the neighbourhood, Definition 3.2.2 can be restated as below.

27
Definition 3.2.4. We say that the limit of a real valued function f (x) is ‘L’ as x approaches a
and write lim f (x) = L if and only if for every neighbourhood N (L) there exists a neighborhood
x→a
Nδ (a) such that x ∈ Nδ (a) ⇒ f (x) ∈ N (L).

3.2.1 Procedure for proving that lim f (x) = L

x→a

The procedure for proving that lim f (x) = L is as follows.

x→a

(1) Choose  > 0, any arbitrary real number.

(2) Find the relationship between |f (x) − L| and |x − a|.
(3) Using the relationship in (2) above, show how we can select a real number δ > 0 such that
|f (x) − L| <  whenever |x − a| < δ (δ depending on ).
Example 3.2.5. Show that lim (2x = 1) = 7.
x→3
Solution: Here f (x) = 2x + 1 and L = 7. Therefore,

f (x) − L| = |2x + 1 − 7| = 2|x − 3|.

So, given  > 0, we then have |2x + 1 − 7| <  provided that |x − 3| < 2 . Therefore, given any
 > 0, we can choose δ() = 2 such that |2x + 1 − 7| <  whenever |x − a| < δ = 2 .

Example 3.2.6. Show that lim x2 = 25.

x→5
Solution: Given any  > 0, we wish to find a δ() such that |x2 − 25| <  whenever 0 <
|x − 5| < δ. Now for some choice of δ, we have that |x − 5| < δ implies that

(3.2.5) |x2 − 25| = |x − 5||x + 5| < δ|x + 5|

If we choose δ 6 1, then

|x − 5| < δ ⇒ |x − 5| < 1 ⇒ 4 < x < 6 and |x + 5| 6 |x| + |5| < 6 + 5 = 11.

Thus, from Equation 3.2.5, |x − 5| < δ ⇒ |x2 − 25| < 11δ whenever δ 6 1. But we wish to have

|x2 − 25| <  which is only possible when δ < 11 . Thus, for a given  > 0, there exists a δ = 2
and δ 6 1.
Example 3.2.7. Show that lim x2 = 16.
x→4
Solution: (As Example 3.2.6).
Theorem 3.2.8. (Uniqueness theorem) If lim f (x) exists then it is unique.
x→a

Proof. Assume that the theorem is false, i.e., there are two different limits of f (x) as x → a.
Let these two limits be L1 and L2 with L1 6= L2 . Then for any given  > 0 there exists a δ1 > 0
such that |f (x) − L1 | < 2 whenever 0 < |x − a| < δ1 . Similarly, for a given  > 0, there exists
a δ2 > 0 such that |f (x) − L1 | < 2 whenever 0 < |x − a| < δ2 . Taking min{δ1 , δ2 } we have
|f (x) − L1 | < 2 whenever 0 < |x − a| < δ and |f (x) − L2 | < 2 whenever 0 < |x − a| < δ. Now
 
(3.2.6) |L1 − L2 | = |L1 − f (x) + f (x) − L2 | 6 |L1 − f (x)| + |L2 − f (x)| < + =
2 2
whenever 0 < |x − a| < δ. So, for any  > 0, we have |L1 − L2 | <  is true only when L1 = L2 .
This is because if L1 6= L2 then |L1 − L2 | = ρ > 0 and for any choice of  6 ρ, the Inequality
(3.2.6) above becomes invalid.

28
 3.2.2 Algebra of limits

(i) Let p(x) = c0 + c1 x + . . . + cn xn , cn 6= 0 be a polynomial of degree n and ‘a’ be a real number.

Then limx→a p(x) = p(a) = c0 + c1 a + . . . + cn an .

Example 3.2.9. Find lim (5x4 + 3x3 − 6x2 − 3x + 5).

x→1
Solution: lim p(x) = p(1) = 5.14 + 3.13 − 6.12 − 3.1 + 5) = 4.
x→1

Example 3.2.10. Find lim c) where c is a constant.

x→a
Solution: Since c is a constant function ( i.e., a polynomial with degree undefined) we get
lim p(x) = p(a) = c.
x→a

Let lim f (x) = `, lim g(x) = m and c be a constant. Then

x→a x→a

(ii) lim (f (x) ± g(x))) = lim f (x) ± lim g(x) = ` ± m.

x→a x→a x→a

(iii) lim cf (x) = c lim f (x) = c`.

x→a x→a

  
(iv) lim (f (x)g(x)) = lim f (x) lim g(x) = `m.
x→a x→a x→a

f (x) lim f (x) `

(v) lim = x→a = for m 6= 0 and g(x) 6= 0 for all x in some neighbourhood of a.
x→a g(x) lim g(x) m
x→a

Proof. We will prove only (iv) and (v).

(iv) First notice that

(3.2.7) |f (x)g(x) − `m| = |f (x)g(x) − f (x)m + f (x)m − `m|

6 |f (x)g(x) − f (x)m| + |f (x)m − `m|
(3.2.8) = |f (x)||g(x) − m| + |m||f (x) − `|.

So given any  > 0, there exists a δ1 > 0 such that |f (x) − `| < 2(|m|+1) whenever

0 < |x − a| < δ1 . Also, there exists a δ2 > 0 such that |g(x) − m| < 2(Mf +mf ) whenever
0 < |x − a| < δ2 where Mf = max |f (x)| and mf = min |f (x)| whenever 0 <
x∈Nδ2 (a) x∈Nδ2 (a)
|x − a| < δ2 . Take δ = min{δ1 , δ2 } then
   
(3.2.9) |f (x)g(x) − `m| 6 Mf + |m| < + = .
2(Mf + mf ) 2(|m| + 1) 2 2

Therefore, given any  > 0 there exists a δ > 0 such that |f (x)g(x) − `m| <  whenever
|0 < |x − a| < δ.
(v) Let  > 0 be given. Then there is some
|m| 1 2
(1) δ1 > 0 such that |g(x)| > 2 or |g(x)| < |m| whenever |x − a| < δ1 .
(2) δ2 > 0 such that |f (x) − `| < |m|
4 whenever |x − a| < δ2 .
|m|2
(3) δ3 > 0 such that |g(x) − m| < 4(|`|+1) whenever |x − a| < δ3 .

29
 Let δ = min{δ1 , δ2 , δ3 } and choose x such that |x − a| < . Then

f (x) ` mf (x) − `g(x) mf (x) − `m + `m − `g(x)

(3.2.10) − = =
g(x) m mg(x) mg(x)
|f (x) − ` |g(x) − m|
6 + |`|
|g(x)| |m||g(x)|
|m| 2 2|`| |m|2  
<  + < + = .
4 |m| |m|2 4(|`| + 1) 2 2

Example 3.2.11. Evaluate the following limits.

x(x2 + 1)
 
(i) lim + x4 + 5 .
x→0 2
√ √
(ii) lim x(1 + x).
x→4

(iii) lim (x5 + 1)3 .

x→−1

x3 + 5x2 + 3x
(iv) lim .
x→−1 2x5 + 5x4 − 2x2 − 1
x2 − 4
(v) lim .
x→2 (x − 2)(x + 3)

3.2.3 Infinite limits and limits at infinity

Recal the notation

(3.2.11) lim f (x) = `

x→a

where a and ` are real numbers means that given any  > 0 there exists a δ > 0 such that
|f (x) − `| <  whenever 0 < |x − a| < δ. The Definition (3.2.11) can be extended to give
definitions of infinite limits and limit at infinity.
Remark 3.2.12. (1) x → ∞: This can be regarded as x approaches ∞ since ∞ is not a real
number. It simply means that x increases without any bounds in the positive direction.
That is, given and M > 0 no matter how large x becomes, it remains bigger than M .
(2) x → −∞: This means that x is negative and |x| grows without any bounds.
(3) lim f (x) = ∞: This means that f (x) grows without bounds in the positive direction as x
x→a
approaches a.
(4) lim f (x) = −∞: This means that f (x) is negative and |f (x)| grows without bounds as
x→a
x approaches a. Now that we have three possibilities each for a and `, i.e., finite, −∞
and +∞: nine (9) combinations are possible for (3.2.11). Below, we define all the nine
possible combinations of (3.2.11) in a tabular form.

30
 Statement of Given any there exists such that whenever
1. lim f (x) = ` >0 δ>0 |f (x) − `| <  0 < |x − a| < δ
x→a
2. lim f (x) = ∞ M >0 δ>0 f (x) > M 0 < |x − a| < δ
x→a
3. lim f (x) = −∞ M <0 δ>0 f (x) < M 0 < |x − a| < δ
x→∞
4. lim f (x) = ` >0 N >0 |f (x) − `| <  x>N
x→∞
5. lim f (x) = ` >0 N <0 |f (x) − `| <  x<N
x→−∞
6. lim f (x) = ` M >0 N >0 f (x) > M x>N
x→a
7. lim f (x) = ∞ M >0 N <0 f (x) > M x>N
x→−∞
8. lim f (x) = −∞ M <0 N <0 f (x) < M x<N
x→−∞
9. lim f (x) = −∞ M <0 N >0 f (x) < M x>N
x→∞

Example 3.2.13. Show that

1
(i) lim = ∞.
x→0 x2
1
(ii) lim 2 = 0.
x→∞ x

Solution:

(i) This is an example of type (2) above where a = 0. We therefore need to show that corre-
spoding to any M > 0 there is a δ > 0 such that f (x) > M whenever 0 < |x − a| < δ, i.e,
0 < |x| < δ. But f (x) > M implies that x12 > M ⇒ M x2 < 1 ⇒ x2 < M 1
or |x| < √1M .
1 1 1
We therefore choose δ = √M . Then 0 < |x| < δ ⇒ |x| < √M ⇒ x2 > M or f (x) > M .
(ii) Given  > 0, we need to find N > 0 such that |f (x) − 0| <  whenever x > N . But, this
implies that |x−2 | <  ⇒ x > √1 . So, choose N = √1 then x > N ⇒ x > √1 ⇒ |f (x)| < 
whenever x > N .
x2 − 7x + 11
Example 3.2.14. Evaluate lim .
x→∞ 3x2 + 10

Solution:
x2 − 7x + 11 1 − x7 + x112 1
lim = lim = .
x→∞ 2
3x + 10 x→∞ 3 + x102 3
1
Theorem 3.2.15. (i) If |f (x)| → ∞ as x → a then |f (x)| → 0 as x → a.
1
(ii) If f (x) → 0 as x → a then |f (x)| → ∞ as x → a.

Theorem 3.2.16. If f (x) → ∞ and g(x) → ` as x → a then:

(i) (f (x) + g(x)) → ∞ as x → a.

(ii) (f (x).g(x)) → ∞ as x → a if ` > 0.
(iii) (f (x).g(x)) → −∞ as x → a if ` < 0.
2 + x2
Example 3.2.17. Evaluate lim .
x→0 | sin(x)|
1
Solution: Let f (x) = | sin(x)| and g(x) = 2 + x2 . Then f (x) → ∞ and g(x) → 2 as x → 0.
2 + x2
Using Theorem 3.2.16 (ii) gives lim = ∞.
x→0 | sin(x)|

31
Theorem 3.2.18. If f (x) → ∞ and g(x) → ∞ as x → a then

(i) (f (x) + g(x)) → ∞ as x → a and

(ii) (f (x).g(x)) → ∞ as x → a.
Remark 3.2.19. Results of Theorems 3.2.16,3.2.18 remain even valid if x → ∞.

3.2.4 One sided limits

Recall that a function f (x) is said to have a limit ` ∈ R at ‘a’ if f (x) approaches ` as x → a
from either sides of a, i.e, from the right of a and from the left of a. There is a possibility
which cannot be ignored, i.e, f (x) approaches ` as x → a from the right (or left) and does not
approach ` as x → a from the left (or right). This type of limit is called a one-sided limit.
Example 3.2.20. Consider the function
(
2x x61
f (x) = .
x2 x>1

As x → 1 from the right, f (x) → 1. Hence, lim+ f (x) = 1. Also as x → 1 from the left,
x→1
lim f (x) = 2. We therefore note the following in this example:
x→1−

(i) The limit of f (x) exists form the right of x = 1.

(ii) The limit of f (x) exists form the left of x = 1.
(iii) The two limits are not the same.
Definition 3.2.21. Let f be a function defined on an open interval containing ‘a’. Then we
say that f (x) has a limit ` ∈ R from the right of a and write lim+ f (x) provided that given any
x→a
 > 0, there exists a δ > 0 (depending on ) such that |f (x) − `| <  whenever a < x < a + δ.
Definition 3.2.22. Let f be a function defined on an open interval containing ‘a’. Then we
say that f (x) has a limit ` ∈ R from the left of a and write lim− f (x) provided that given any
x→a
 > 0, there exists a δ > 0 (depending on ) such that |f (x) − `| <  whenever a − δ < x < a.
Example 3.2.23. Show that
√
(i) lim x = 0.
x→0+
√
(ii) lim− x does not exist.
x→0

Solution:

√0 be given. We√need to find a δ() > 0 such that |f (x)−0| <  whenever
(i) Let  > √0 < |x| < δ.
Now | x − 0| <  ⇒ x <  ⇒ x < 2 . So √ an obvious choice of δ is 2
so | x − 0| < 
whenever 0 < x < δ = 2 . Therefore, lim+ x = 0.
x→0
√
(ii) Since x does not remain a real√valued function when x < 0, it is not defined for values
to the left of zero and thus lim x = 0 does not exist.
x→0−

Remark 3.2.24. Definitions of one sided limits are also similar to Definitions 3.2.21, 3.2.22.
A careful observation of the difference btween Definition 3.2.2 and Definitions 3.2.21, 3.2.22
can guide on how to write the definitions of one sided infinite limits from the table in Subsection
3.2.3.

32
Theorem 3.2.25. lim f (x) = ` if and only if lim f (x) = ` and lim f (x) = `.
x→a x→a+ x→a−

Proof. Let us first establish that lim f (x) = ` if and only if lim f (x) = ` and lim f (x) = `.
x→a x→a+ x→a−
Now suppose that lim f (x) = `. Then given  > 0 there exists a δ > 0 such that |f (x) − `| < 
x→a
whenever 0 < |x − a| < δ ⇒ |f (x) − `| <  whenever a − δ < x < a + δ ⇒ |f (x) − `| <  whenever
a − δ < x < a and |f (x) − `| <  whenever a < x < a + δ ⇒ lim+ f (x) = ` and lim− f (x) = `.
x→a x→a

Now lim+ f (x) = ` implies that given any  > 0 there exists a δ1 > 0 such that |f (x) − `| < 
x→a
whenever a < x < a + δ1 and lim f (x) = ` implies implies that given any  > 0 there exists a
x→a−
δ2 > 0 such that |f (x) − `| <  whenever a − δ2 < x < a. Therefore, take δ = min{δ1 , δ2 } then
we have  (
 lim f (x) = ` |f (x) − `| <  whenever
x→a +
⇒
 lim f (x) = `
−
a−δ <x<a+δ
x→a

Therefore, we have lim f (x) = `.

x→a

Remark 3.2.26. (1) Theorem 3.2.25 is also valid for infinite limits and we have that lim f (x) =
x→a
∞ implies that lim+ f (x) = ∞ and lim− f (x) = ∞.
x→a x→a
(2) Results analogous to those given in §3.2.2 also holds for one sided limits.
Example 3.2.27. Consider the absolute value function
(
x x>0
f (x) = |x| = .
−x x < 0

We have that lim+ |x| = lim+ x = 0 and lim− |x| = lim− −x = 0. Since lim+ |x| = lim− |x| =
x→0 x→0 x→0 x→0 x→0 x→0
0 then by Theorem 3.2.25, lim |x| = 0.
x→0

Theorem 3.2.28. (Squeezing theorem). If f, g and h are functions defined on an interval and
satisfying g(x) 6 f (x) 6 h(x) on R, and lim g(x) = lim h(x) = `, then limx→a f (x) = `.
x→a x→a

Proof. lim g(x) = ` implies that given any  there exists δ1 > 0 such that |g(x) − `| < 
x→a
whenever 0 < |x − a| < δ1 ⇒ ` −  < g(x) < ` +  whenever 0 < |x − a| < δ1 . Also,
lim h(x) = ` implies that given any  there exists δ2 > 0 such that |h(x) − `| <  whenever
x→a
0 < |x − a| < δ2 ⇒ ` −  < h(x) < ` +  whenever 0 < |x − a| < δ2 . Now
 (
 lim g(x) = ` ` −  < g(x) 6 f (x) 6 h(x) < ` + 
x→a ⇒ .
 lim h(x) = ` whenever 0 < |x − a| < δ
x→a

This implies ` −  < f (x) < ` +  whenever 0 < |x − a| < δ. Hence, lim f (x) = `.
x→a

Example 3.2.29. Exanine the limiting behaviour of the function f (x) at x = 2 given that
2x2 + 3x + 18 6 f (x) 6 4x3 .
Solution: Here f (x) is squeezed between two functions 2x2 + 3x + 18 and 4x3 . Now lim 2x2 +
x→2
3x + 18 = 32 and lim 4x3 = 32. Hence, f (x) satisfies the condition of Theorem 3.2.28 so that
x→2
lim f (x) = 32.
x→2

33
Remark 3.2.30. One of the very interesting applications of the squeezing theorem lies in the
evaluation of the function sin(x)
x at point x = 0. Here sin(x)
x for 0 6 x 6 π2 can be squeezed
1 1 sin(x)
between the two functions cos(x) and cos(x), i.e., cos(x) 6 x 6 cos(x).
sin(x)
Example 3.2.31. Find lim .
x→0 x
1
Solution: Since lim = 1 and lim cos(x) = 1 then by squeezing theorem, we conclude
x→0 cos(x) x→0
sin(x)
that lim = 0.
x→0 x

3.3 Continuity

Every function can be represented by its graph. We can therefore consider the graph of a function
as one of the ways that gives the idea of continuous functions. Consider the graphs (1) and (2)
below.

We observe that in graph (1), the curve AB has no break but in graph (2), the curve is broken
at x = a as seen from the two pieces Ac and DB. Therefore, the function represented by the
graph (1) is said to be continuous for all values in its domain whereas the function represented
by the graph (2) is not continuous, i.e discontinuous at x = a.
Definition 3.3.1. The function f is said to be continuous at a if the following three (3)
conditions are satisfied:

(i) f is defined at a, i.e., a ∈ D(f ).

(ii) f has a limit at a i.e., lim f (x) exists.
x→a
(iii) lim f (x) = f (a).
x→a
Remark 3.3.2. If a function is continuous for every x ∈ R, then it is said to be an everywhere
continuous function.
Example 3.3.3. Discuss the continuity of the function f (x) = x2 in R.
Solution: Let a be an arbirtary real number. We observe that

(i) f (a) = a2 .
(ii) lim f (x) = lim x2 = a2 .
x→a x→a
(iii) lim f (x) = f (a).
x→a

Therefore, all the three conditions in Definition 3.3.1 are satisfied showing that f (x) is con-
tinuous at x = a. But a ∈ R is arbitrary, hence, it follows that f (x) = x2 is an everywhere
continuous function.
Example 3.3.4. Discuss the continuity of the absolute valued function
(
x x>0
f (x) = |x| = .
−x x < 0

34
Solution:
For x > 0: Take any real number a > 0. Then lim f (x) = lim |x| = a = f (a), i.e., lim f (x) =
x→a x→a x→a
f (a). Thus the function is continuous for all positive real values of x.

For x < 0: Take any real number b < 0. Then lim f (x) = lim |x| = −b = |b| = f (b), i.e.,
x→b x→b
lim f (x) = f (b). Thus the function is continuous for all negative real values of x.
x→b

For x = 0: We have that f (0) = 0 and lim f (x) = lim |x| = 0, i.e., lim f (x) = f (0). Thus
x→0 x→0 x→0
the function is continuous at x = 0.

So, it follows that the absolute value function is an averywhere continuous fucntion on R.
Example 3.3.5. Discuss the continuity of the function
(
2x − 1 x < 1
f (x) =
x x>1

at x = 1.
Solution:
Observe that lim f (x) = lim x = 1 and lim f (x) = lim 2x − 1 = 1. So, lim f (x) = 1
x→1+ x→1+ x→1− x→1− x→1
( i.e., lim f (x) exists). Further, since f (1) = 1 we get that lim f (x) = f (1). Hence, f (x) is
x→1 x→1
continuous at x = 1.
Definition 3.3.6. The function f (x) is said to be continuous over an open interval (a, b) if it
is continuous at every point of the interval.
Remark 3.3.7. Note that the endpoints of the open interval a and b in Definition 3.3.6 may
also be taken as ∞ and ∞. So, this definition is valid for intervals (∞, b), (a, ∞ and (−∞, ∞).
Functions in the three examples above are continuous in the open intervals (−∞, ∞).

3.3.1 Formal Definition of continuity

A function f (x) is said to be continuous at a ∈ D(f ) if given any  > 0, there exists a δ > 0
such that

(3.3.1) |f (x) − f (a)| <  whenever 0 6 |x − a| < δ.

Remark 3.3.8. Note the difference in inequality (3.3.1) and the corresponding inequality in
the case of limit of f at a. The inequality (3.3.1) emphasizes that in case of continuity, values
of x are not considered in the deleted neighbourhood of a, i.e., 0 < |x − a| < δ, as in the case
of limits. Reason being that continuity at a point necessitates that the function f be defined at
that point.
Example 3.3.9. Discuss the continuity of the greatest integer function [x] at x = 3 where for
a fixed integer x, the greatest integer function [x] = n where n 6 x < n + 1.

Solution: Observe that the given function is defined at x = 3 and its value f (3) = [3] = 3.

The existence of lim [x] can be seen by considering the limit from the right and left of x = 3.
x→3
Consider the limit from the right; the value of x can be considered in the range 3 6 x < 4 and
by definition in this range [x] = 3, i.e., lim+ [x] = 3. Further, consider the range 2 6 x < 3.
x→3
We then obtain lim− [x] = 2.
x→3

35
Therefore, lim [x] 6= lim [x]. Thus, the limit of the function does not exist at x = 3. This
x→3+ x→3−
violets Condition of Definition 3.3.1 (ii). Hence, the greatest integer function is not continuous
at x = 3.

By the same reasoning, as in this example, it can be concluded that the gretest integer function
is not continuous at any integral value of x.
Remark 3.3.10. One of the interesting observations is the fact that the function [x] is not
continuous at x = 3 but its limit from the right at x = 3 is equal to the value of x = 3, i.e.,
lim [x] = [3] = 3 = f (3). In this case, we can say that the function is continuous from the
x→3+
right at x = 3.
Definition 3.3.11. A function f is said to be continuous at a from the right if lim f (x) =
x→a+
f (a). It is said to be continous from the left if lim− f (x) = f (a).
x→a

Remark 3.3.12. The type of continuity in Definition 3.3.11 is called one sided continuity.
However, a function f is continuous at a if it is continous from the left and right of a.
Theorem 3.3.13. (Theorems on continuity). Let f (x) and g(x) be continuous functions at
x = a and c ∈ R. Then

(i) the sum of f (x) and g(x) is continuous at x = a.

(ii) cf (x) is continuous at x = a.
(iii) the product f (x).g(x) is continuous at x = a.
f (x)
(iv) if g(a) 6= 0, the quotient g(x) is continuous at x = a.
(v) the identity function f (x) = x and the constant function g(x) = c are continuous for all
values of x.
(vi) any polynomial f (x) = a0 + a1 x + . . . + an xn an 6= 0 is a continuous function for all values
of x.
(vii) if R(x) = fg(x)
(x)
is a rational function where f (x) and g(x) are polynomials then R(x) is
continuous for all those vaues of x for which g(x) 6= 0.
Theorem 3.3.14. Let g be a continuous function at a and f be a continuous at g(a). Then
the composite function f ◦ g is continuous at x = a.

Proof. Note that

(i) The continuity of f at g(a) implies that given any  > 0, there exists a δ1 > 0 such that
|f (g(x)) − f (g(a))| <  whenever 0 6 |g(x) − g(a)| < δ1 .
(ii) The continuity of g at a implies that given any 1 > 0, there exists a δ > 0 such that
|g(x) − g(a)| < 1 whenever 0 6 |x − a| < δ. On choosing 1 = δ1 , which can be done
in view of the fact 1 can be chosen at our wish, we have that |g(x) − g(a)| < δ1 = 1
whenever 0 6 |x − a| < δ.
(iii) Combining (i) and (ii) we obtain |g(x) − g(a)| < δ1 whenever 0 6 |x − a| < δ. But
f (g(x)) − f (g(a))| <  whenever 0 6 |g(x) − g(a)| < δ1 . Hence, f (g(x)) − f (g(a))| < 
whenever 0 6 |x − a| < δ. In otherwords, given any  > 0 there exists a δ > 0 such that
f (g(x)) − f (g(a))| <  whenever 0 6 |x − a| < δ. This implies that f ◦ g is continuous at
x = a.

36
Example 3.3.15. Consider the function f (x) = | sin(x) + cos(x)|. Justify that

lim | sin(x) + cos(x)| = | sin(π) + cos(π)| = 1.

x→π

Solution: The above expression implies that lim f (x) = f (π). This can be justified if we show
x→π
that f (x) is continuous at x = π. Let h(x) = sin(x) + cos(x) and g(x) = |x|. Hence h(x)
and g(x) are continuous at x = π (Show this!). Then by Theorem 3.3.14, we have that the
composite function g(h(x)) is also continuous at x = π. But g(h(x)) = g(sin(x) + cos(x)) =
| sin(x) + cos(x)| = f (x). We therefore conclude that f (x) = | sin(x) + cos(x)| is continuous at
x = π and lim | sin(x) + cos(x)| = | sin(π) + cos(π)| = 1.
x→π

Remark 3.3.16. Note that

(i) having knowledge of continuity of some elementary functions, Theorem 3.3.14 can be used
for predicting the continuity of some complicated functions without actually using the def-
inition of continuity. This is seen in the above example, i.e., to establish the continuity of
a given function f then g and h must be suitably chosen.
(ii) statements such as ‘f is a continuous function’, ‘the function is continuous’, where the
point at which it is continuous is not specified means that the function is continuous for
all values of its domain.
√
Example 3.3.17. Show that the function f (x) given by f (x) = x4 + 5 is continous.
Solution: The function should be written as the composite function of two functions g(x) and √
h(x) whose continuity is already known or can be established. Consider the functions g(x) = x
and h(x) = x4 + 5. Then
p
(g ◦ h)(x) = g(h(x)) = g(x4 + 5) = x4 + 5 = f (x).

But g(x) and h(x) are both continuous functions. Therefore, by Theorem 3.3.14, it follows that
f (x) is continuous.
Example 3.3.18. Determine the values of a, b and c for which the function
 sin((a+1)x)+sin(x)

 x x<0
f (x) = c x=0
 (x+b2 )1/2 −x1/2

bx3/2
x>0

is continuous at x = 0.
Solution: It is given the function f is continuous at x = 0 and f (0) = c. Now the existence
of lim f (x) can be seen from the left and from the right of x = 0. That is,
x→0

(x + bx2 )1/2 − x1/2 (1 + bx1/2 ) − 1

(3.3.2) lim+ f (x) = lim+ = lim
x→0 x→0 bx1/2 x→0+ bx
b2
(1 + 2b x − 8 + . . .) − 1 1
= lim = ,
x→0+ bx 2
and
 
sin(a + 1)x + sin x (a + 1) sin(a + 1)x sin(x)
(3.3.4) lim− f (x) = lim− = lim− + = a+2.
x→0 x→0 x x→0 (a + 1)x x

For continuity of the function at x = 0, lim f (x) = lim f (x) = f (0). So, we must have
x→0+ x→0−
1 3 1
a + 2 = = c, i.e, a =
2 c= 2, 2.
Further b can be any real number except 0, because in this
case the function would not be possibly defined for x > 0.

37
 Definition 3.3.19. A function f is said to be continuous on a closed interval a 6 x 6 b if it
satisfies the following three conditions.

(i) lim f (x) = f (c), a < c < b.

x→c
(ii) lim f (x) = f (a).
x→a+
(iii) lim− f (x) = f (b).
x→b

Theorem 3.3.20. (The intermediate value property (IVP)). If f is a real valued function
continuous on [a, b] and K is a real number between f (a) and f (b) then there is at least one
number c between a and b such that f (c) = K. Consider the two graphs below

The graph in figure (b) is discontinuous at x = c and it can be seen that for some K lying
between f (a) abd f (b) in figure (a) above there exists a c between a and b such that f (c) = K
while in figure (b) such a value does not exist.

3.3.2 Discontinuity

Recall that the function f is said to be continuous at a if

(i) f is defined at a.
(ii) lim f (x) exists.
x→a
(iii) lim f (x) = f (a).
x→a

The function which is not continuous at a is said to be discontinuous at a and the point a is the
point of discontituity of the function. The function is therefore discontinuous at x = a if any of
the above three conditions is voilated and such a discontinuity is called simple discontinuity.
It can arise from any of the following.

(1) Lack of definition of the function at x = a: A function which is not defined at a

violates the condition (i) and hence it is discontinous at a. This type of discontinuity is divided
into two categories.

(a) Limit of f at a exists: The discontinuity of this type can be removed (hence called a
removable discontinuity) by extending the definition of f to include its limit at x = a. The
new function is called the continuous extension of f .
Example 3.3.21. Test its continuity of the function f (x) given by f (x) = sin(x)
x at x = 0.
sin(x)
Solution: Notice f (x) is not defined at x = 0 but lim = 1. So, f (x) is discontin-
x→0 x
uous at x = 0. Now redifine f to give a new function F (x) given by
(
sin(x)
x x 6= 0
F (x) =
1 x=0

38
 gives an everywhere continuous extension of f (x). (Note: This is the case of a removable
discontinuity).
(b) Limit of f at a does not exist: In general, functions of this nature show an infinite
behaviour such as |f (x)| → ∞ as x → a ad such a discontinuity cannot be removed. This
type of dscontinuity is called infinite or irremovable discontinuity.
1
Example 3.3.22. Test the continuity of the function f (x) = x−1 at x = 1.
Solution: We observe the following
(i) f (1) does not exist
1 1
(ii) lim− f (x) = lim f (1 − h) = lim = −∞, lim+ f (x) = lim f (1 + h) = lim = ∞.
x→1 h→0 h→0 −h x→1 h→0 h→0 h
Therefore, we see that f (x) has an infinite discontinuity at x = 1. This cannot be removed.

(2) Failure of the limit to exist at a: If the limit of the function does not exist at a, then it
violates condition (ii) of the definiton of continuity at th point a and hence the function becomes
discontinuous at that point a. We also have two types of this case.

(a) The function defined at a: The limit of the function at point a fails to exist if
(i) lim+ f (x) and lim− f (x) are both finite but not equal. In such a case the function is
x→a x→a
said to have a Jump discontinuity at x = a.
(ii) Only one of the limits lim+ f (x) of lim− f (x) exists. In such a case the function is
x→a x→a
said to have a mixed discontinuity at x = a.
(iii) Neither lim+ f (x) nor lim− f (x) exists.
x→a x→a
(b) Function not defined at a: This case coincides with the case 1(b) since in both cases
neither the limits exist at a nor the function is defined at a.

(3) Failure of limit and value of the function to agree at a: Functions of this type are
discontinuous at a because they violate condition (iii) of the definition of continuity.
Example 3.3.23. Test the continuity of the function
( 2
x −1
x 6= 1
f (x) = x−1
0 x=1

at x = 1.
Solution: It can be seen that lim f (x) = 2 and f (1) = 0. Therefore, lim f (x) 6= f (1). Hence,
x→1 x→1
the function is discontinuous at x = 1. However, the discontinuity can be removed by redefining
the function as ( 2
x −1
x 6= 1
f (x) = x−1 .
2 x=1

Definition 3.3.24. The function is called piecewise continuous on an interval (a, b) if its
continuous in (a, b) except for at most a finite number of jump discontinuities.
Diagram

39
Chapter 4

Differentiation and continuity

4.1 Differentiation

Definition 4.1.1. Let f (x) be a continuous function defined on an open interval (a, b). Let
x0 ∈ (a, b). Then, the derivative of f (x) at x0 with respect to x, denoted by f 0 (x) is defined as

f (x) − f (x0 )
f 0 (x) = lim .
x→x0 x − x0

If the limit of the above quotient on the right hand side exists, we say that the function f (x) is
differentiable at x = x0 . If we set x = x0 + h, then the above equation becomes

f (x) − f (x0 )
f 0 (x) = lim .
h→0 h
Example 4.1.2. Show that the derivative of f (x) = x2 exists at any value x0 ∈ R and finite.
Solution
f (x) − f (x0 ) x2 − x20
f 0 (x0 ) = lim = lim = lim x + x0 = 2x0 .
x→x0 x − x0 x→x0 x − x0 x→x0

Example 4.1.3. Show that the function f (x) defined by

(
x + tan−1 ( x1 ), x 6= 0
f (x) = .
0, x = 0.

is continuous at x = o but not differentiable there.

Solution
Continuity at x = 0: To establish the continuity at x = 0, we need to check whether

lim f (x) = lim− f (x) = f (0).

x→0+ x→0

Now  
−1 1 π
lim+ f (x) = lim f (0 + δx) = lim (0 + δx) tan =0× =0
x→0 δx→0 δx→0 0 + δx 2
and  
−1 1 π
lim f (x) = lim f (0 + δx) = lim (0 + δx) tan =0× = 0.
x→0− δx→0 δx→0 0 + δx 2

40
But f (0) = 0, therefore, the given function is continuous at x = 0.
Differentiability at x = 0: If the function is differentiable at x = 0 then its derivative at that
point is given by
f (0 + h) − f (0)
f 0 (0) = lim .
h→0 h
To check the existence of the derivative of f (x) at x = 0, we need to check whether Rf 0 (0) =
Lf 0 (0). That is,

h tan−1 h1 − 0


0 f (h) − f (0) π
Rf (0) = lim = lim =
h→0 + h h→0 h 2
and  
f (h) − f (0) −h tan−1 −h
1
−0 π
0
Lf (0) = lim = lim =− .
h→0 − −h h→0 −h 2
So, it is easily seen that Rf 0 (0) 6= Lf 0 (0) and so f (x) is not differentiable at x = 0.
Exercise 4.1.4. Test the differntiability of the function
(
x2 sin x1


x 6= 0
f (x) =
0 x=0

at x = 0.

4.1.1 Continuity and Differentiability

Continuity of the function f (x) at x0 is necessary for existence of f 0 (x) at x0 .

Assume that f (x) is not continuous at x0 . This implies that either one or more of the following
statements are true
[(i)]f (x0 ) does not exist. limx→x0 f (x) does not exist. limx→x0 f (x) 6= f (x0 ).

It is now a simple matter to verify that if any of the above statement is true then the expression
f 0 (x) = limx→x0 f (x)−f
x−x0
(x0 )
becomes meaningless and hence f 0 (x0 ) does not exist.

Continuity at x0 is not sufficient for existence of the derivative at x0

We justify the above statement by giving the function and the point in its domain where the
function is continuous but the derivative does not exist.
ExampleConsider the function f (x) = |x| at x = 0. This function is continuous everywhere in
its domain R and therefore continuous at x = 0. The derivative of the given function at x = 0
is given by (
0 f (x) − f (x0 ) |x| 1 x≥0
f (0) = lim = lim =
x→x0 x − x0 x→0 x −1 x = 0.
Hence the derivative of a function f (x) = |x| does not exist at x = 0 since Rf 0 (0) 6= Lf 0 (0).

41
Differentiability of a function f (x) at x0 implies continuity of a function at x0

Justification; Let the derivative of the function f (x) exist at x0 . This implies that Rf 0 (x) =
Lf 0 (x). So it is enough to show that

lim f (x) = f (x0 ).

x→x0

This is done below

f (x) − f (x0 ) f (x) − f (x0 )
lim (f (x) − f (x0 )) = lim (x − x0 ) = lim lim (x − x0 ) = 0.
x→x0 x→x0 x − x0 x→x0 x − x0 x→x0

Hence limx→x0 f (x) = f (x0 ) and theefore differentiability implies continuity.

4.1.2 Theorem (Algebra of Derivatives)

Let f (x) and g(x) be continuous functions on a common interval (a, b). If the derivative of f (x)
and g(x) both exist at x0 in (a, b) then
[(i)]
[f + / − g]0 (x0 ) = f 0 (x0 ) + / − g 0 (x0 )

[f × g]0 (x0 ) = f 0 (x0 ) × g(x0 ) + f (x0 )g 0 (x0 )

f  g(x0 ) × f 0 (x0 ) − g 0 (x0 )f (x0 )

(x0 ) = , for g(x) 6= 0.
g [g(x0 )]2

Proof
f (x)−f (x0 ) g(x)−g(x0 )
Define f 0 (0) = limx→x0 x−x0 and g 0 (0) = limx→x0 x−x0 .
[(i)]
[f (x)+g(x)]−[f (x0 )+g(x0 )
[f + g]0 (x0 ) = limx→x0 x−x0 ,
f (x)−f (x0 ) g(x)−g(x0 )
= limx→x0 x−x0 + limx→x0 x−x0 ,
= f 0 (x0 ) + g 0 (x0 ).

Exercise
f (x) f (x0 )
f  g(x) − g(x0 )
(x0 ) = lim
g x→x0 x − x0
f (x)g(x0 )−g(x0 )f (x0 )
g(x)g(x0 )
= lim
x→x0 x − x0
f (x)g(x0 ) − f (x0 )g(x0 ) + f (x0 )g(x0 ) − g(x)f (x0 )
= lim
x→x0 g(x)g(x0 )(x − x0 )
f (x) − f (x0 )
1 g(x) − g(x0 )
g(x0 )
= lim × − lim ×
x→x0 x − x0 g(x) x→x0 x − x0 g(x)g(x0 )
g(x0 ) × f 0 (x0 ) − g 0 (x0 )f (x0 )
=
[g(x0 )]2

42
Corollary

If one of the functions, say g(x), is a constant function, then g(x) = c, implies that g 0 (x) = 0.
With this particular choice of one function, part (ii) of the above theorem yields the following
result;
[c × f ]0 (x0 ) = f 0 (x0 ) × g(x0 ) + f (x0 )g 0 (x0 ) = 0 × f (x0 ) + f 0 (x0 ) × c = cf 0 (x0 )

Example

Establish the differentiability of the function f (x) = √1x in the domain x > 0, and find its
derivative. Solution Let x0 be an arbitrary point in the domain x > 0, then
f (x0 + h) − f (x0 ) 1 1

Rf 0 (x0 ) = lim = lim √ −√
h→0 h h→0 x0 + h x0
1  −1 h −1 −1 
= lim x02 (1 + ) 2 − x 2
h→0 h x0
−1
x2  h
= lim 0 1 −

+ ... − 1
h→0 h 2x0
−1
x2  h
= lim 0

− + higher order terms in h
h→0 h 2x0
1
=− 1
2x02
Similary,
f (x0 + h) − f (x0 ) 1 1

Lf 0 (x0 ) = lim = lim √ −√
h→0 −h h→0 x0 + h x0

1  −1 h −1 −1 
= lim − x 2 (1 − ) 2 − x 2
h→0 h 0 x0
−1
x2  h
= lim − 0 1 −

+ ... − 1
h→0 h 2x0
−1
x02  h 
= lim − + higher order terms in h
h→0 h 2x0
1
=− 1
2x02
Thus, the function is differentiable at x0 since Rf 0 (x0 ) = Lf 0 (x0 ). Further since x0 is arbitrary
point in the domain x0, it follows that the given function is differentiable in the whole domain
and its derivative is f 0 (x0 ) = − 1 1 .
2x02

Questions

3. Compute the derivative, f 0 (x) of the function f (x) = sin x at any point x.
2.
1.
2. Using definition of derivative of a function at a point, show that f 0 (x) = 1
x where f (x) =
loge x.
3. From the definition of the derivative obtain f 0 (x) where f (x) = ex .
4. Obtain f 0 (x) where f (x) = sin−1 x.

43
Chapter 5

Differentials

5.0.3 Introduction

Let y = f (x) be a given function. If δy is the increment iny corresponding to an infinite small
( a quantity whose limit is zero) increment in δx in x, then the quotient

δy f (x + δx) − f (x)
= lim ,
δx δx→0 δx
is ultimately very close to the derivative of y = f (x) given by the expression

dy f (x + δx) − f (x)
= lim .
dx δx→0 δx
dy δy
If  is some arbitrary small number, then the closeness in the value of dx , δx can be interpreted
as
δy dy
− =
δx dx
dy
δy = δx + δx
dx
δy = f 0 (x)δx + δx

5.0.4 Definition

In the expression above δx is called the differential of x and the product 0 (x)δx is known as the
differential of the function y at x with increment δx.
Te differential of the function y = f (x) at x with increment δx is denoted by df (x, δx) or just
dy.
Note
[(i)]The differential df (x, δx) of the function f (x) depends upon a point x and the incre-
ment δx, so it is a function of two variables x and δx. The term δx is a product of
two arbitrary small quantities  and δx which can be ignored as δx → 0. The method of
differentials can be used successfully in evaluating the derivative of a given function. This
can be done as follows; Given the function f (x), first calculate its differential df (x, δx),
then the coefficient of δx in df (x, δx) is the required derivative of f (x) with respect to x.

44
5.0.5 Examples
3. Using the method of differential find the derivatives of the functions
2.
1.
[(i)]
(a) y = x3 ,
(b) y = x1 .
solutions
[(i)]

δy =f (x + δx) − f (x),
=(x + δx)3 − x3 ,
=x3 + 3x2 δx + 3x(δx)2 + (δx)3 − x3 ,
=3x2 δx + δx, where  = (3xδx + (δx)2 ).

Since δx is of order higher than one in δx, then it can be ignored and the differential
of y at x is given by dy = 3x2 δx. Thus f 0 (x) = 3x2 .

δy =f (x + δx) − f (x),
=(x + δx)−1 − x−1 ,
δ
−1
= − 2 x(x + δx) ,
x
δx δx
−1
=− 2 1+ ,
x x
δx δx (δx)2

=− 2 1− + − ... ,
x x x2
δx (δx)2 (δx)3
=− 2 − 3
+ − ...,
x x x4
δx
= − 2 + δx.
x
So, in view we have f 0 (x) = − x12
Remarks; The concept of differential is approximating calculations and estimating errors
in formulae.
(a)
(b)
2. Use differentials to find the approximate value of
√ 1
[(i)] 98 (1010) 3 tan 47
solutions
√ √
[(i)]First observe 98 is closed to 100 and 100 = 10 (known). Take y = x. We
now obtain δy as x changes from 100 t0 98. Thus, for x = 100, δx = −2 and
(−2)
δy ≈ 2δx
√ = √ = −1
x
√ 2 100√ 10 1
Therefore 98 ≈ 100+δy = 10−0.1 = 9.9 Here 1010 is closed to 1000 and (1000) 3
is known to be 10.
Thus as x changes from 1000 to 1010 we have δx = 10, δy ≈ dy = f 0 (x)δx at
x = 1000.
Therefore,
δx 10 1
δy ≈ 2 = 2 = = 0.033
3x 3 3(1000) 3 30
1 1
Thus (1010) 3 ≈ (1000) 3 + δy = 10 + 0.033 = 10.033 Choose the function y = tan x.
This function and its derivative are known at x = 45 so that at 45, y = tan 45 = 1

45
 and y 0 (45) = sec2 45 = 2.
Note that 45 is closed to 47 and therefore we take δ = 2 ≈ 0.035rad Then

δy ≈ dy =y 0 δ at x = 45
= sec2 xδx
=2(0.035)
=0.07.

It now follows that tan 47 ≈ tan 45 + δy = 1 + 0.07 = 1.07

5.0.6 Rolle’s Theorem

Let f (x) be a function continuous on the closed interval [a, b] and differentiable on the interval
(a, b). If f (a) = f (b), then there exists at least one value c in (a, b) such that f 0 (c) = 0.
Proof
If f (x) = c, a constant on [a, b] then obviously f 0 (x) = 0 for all x ∈ (a, b). If f (x) is not a
constant on [a, b] then there are three possibilities, namely:
[(a)]f (x) > f (a) for all x ∈ (a, b). f (x) < f (a) for all x ∈ (a, b). f (x) > f (a) for some
values of x ∈ (a, b) and f (x) < f (a) for the rest.

In all these situations, there exist at least one point p on the curve, tangent at which it is parallel
to the x-axis. If x-coordinate of p then it follows that f 0 (c) = 0

Example

Consider the function f (x) = x3 − 3x2 on the interval [0, 3].

Since f (x) is a polynomial in x, it is continuous on [0,3] and differentiable on (0,3).
Also f (0) = f (3) = 0. Hence all the three conditions in Rolle’s theorem are satisfied.
This guarantees the existence of at least one point in the open inerval (0,3) at which the deriva-
tive is zero.
f 0 (x) is given by
f 0 (x) = 3x2 − 6x.
Therefore, f 0 (x) = 0; 3x2 − 6x = 0; x = 0 or2. But 0 6= (0, 3) and so c = 2 is the point whose
existence is guaranteed by Rolle’s theorem.

46
Failure of any one of the three conditions of Rolle’s theorem to be satisfied

[(a)]If f (a) 6= f (b);If the condition f (a) = f (b) is removed, the point c at which the
derivative is zero may or may not exist Example

Consider the function f (x) = (x − 1)(4x − 3) on the interval [1,4]. The given function
is continuous on the closed interval [1, 4] and differentiable on the open interval (1,4).
However since f (1) = 0 6= f (4) = 39. Let us now find c if it exists at which the derivative
f 0 (x) = 0.
This implies that 4x − 3) + (4x − 4) = 0; x = 87 . But 78 does not lie in the interval
(1,4). This justifiesthe assertion that if the condition f (a) = f (b) is removed, then there
may or may not exist any point c ∈ (a, b) at which the derivative is zero. If f (x) is not
differentiable at x0 ∈ (a, b).
I the condition of differentiability is removed, then again there may not exist any point in
the open interval (a,b) at whih the derivative is zero. Example

onsider the function f (x) = |x| on the interval [-1,1]. The given function satisfies
the following conditions;[(i)]
(a)
(c)
(b)
3. (a) f (x) is continuous on the closed interval [-1,1].
2.
1.
(b) f (−1) = f (1) = 1.
(c) f (x) is not differentiable on the open interval (-1,1) since its derivative doesn’t exist
at x = 0.
Infact; 
1
 x ∈ (0, 1),
f 0 (x) = −1 x ∈ (−1, 0),

does not exist x = 0.


Showing that at no point in the open interval (-1,1) the derivative of the function is
zero.
3. f (x) is not continuous on the closed interval [a,b].
In this case, there are two possibilities;
[(i)]If f (x) is not ontinuous on (a,b) then it can not be differentiable on an open
interval (a,b) and hence in view of (2), Rolle’s theorem fails. If f (x) is continuous
in the open interval (a,b) but fails to be continuous at any of the end points then also
there does not exist a point in the open interval (a,b) where the derivative of f (x)
vanishes. Example below illustrate this point
Example Consider the function
(
x 0 ≤ x ≤ 1,
f 0 (x) =
0 x = 1.

This function has the following properties[(i)]

(a)
(b) i. f (x) is continuous on the open interval (0,1).
ii. f (x) is differentiable on the open interval (0,1)
iii. f (0) = f (1) = 0 but f (x) is not continuous on the closed interval [0,1] sine it
is discontinuous at x = 1. Thus there does not exist any point c in the open
interval (0,1) such that f 0 (c) vanishes there. Rolle’s theorem therefore fails.

More Examples

1. Show that the function f (x) = x2 − 3x + 6 satisfies all the requirements of Rolle’s theorem
on the interval [1,2]. Find the point at which the derivative is zero. Solution

47
 Since the given function is a polynomial of degree 2 in x, it is continuous on the closed
interval [1,2] and differentiable on the open interval (1,2). Also f (1) = 4 = f (2). Thus
all the requirements of Rolle’s theorem are satisfied. There must exist a point at which
the derivative is zero. So, f 0 (x) = 0 = 2x − 3 ⇒ x = 23 and it is this value of x that lies
in the interval (1,2) that Rolle’s theorem gurantees.
2. If f (x) and g(x) are differentiable functions for all 0 ≤ x ≤ 1 such that f (0) = 2,g(0) = 0
and f (1) = 6, g(1) = 2 then show that there exists a c satisfying 0 < c < 1 and f 0 (c) =
2g 0 (c) Solution
In this case we construct a function

F (x) = f (x) − 2g(x),

that satisfies the conditions of Rolle’s theorem on the interval [0,1]. To verify that F (x)
satisfies all the conditions of Rolle’s theorm on the closed interval [0,1] first observe that
since f (x) and g(x) are differentiable on [0,1], they are necessarily continuouw on the
closed interval [0,1]. Now
[(i)]Continuity of f (x) and g(x) on [0,1] implies continuity of F (x) on [0,1]. Differentiability
of f (x) and g(x) on [0,1], implies the the differentiability of F (x) on [0,1]. F (0) =
f (0) − 2g(0) = 2 and F (1) = f (1) − 2g(1) = 2. Hence F (0) = F (1)
Therefore all the conditions of Rolle’s theorem are satisfied by the functon F (x) on
[0,1]. Hence there must exist a value c ∈ (0, 1) such that F 0 (c) = 0.
But F (x) = f (x) − 2g(x). Therefore

F 0 (x) = f 0 (x) − 2g 0 (x).

Hence F 0 (c) = 0 ⇒ f 0 () − 2g 0 (c) = 0 so that f 0 (c) = 2g 0 (c).

Consquences of Rolle’s Theorem

Result 1: Let f (x) be differentiable in the open interval (a,b). Then between any two zeros of
f (x), witin this interval, there lies a ero of f 0 (x). Proof
Let x1 and x2 be two zeros of f (x) with in the interval [a,b). It follows that f (x1 ) = f (x2 )
a < x1 < b. Further, the differentiability of f (x) on (a,b) makes
[(i)]f (x) continuous on [x1 , x2 ], f (x) differentiable on [x1 , x2 ] and f (x1 ) = f (x2 ).

Thus the conditions of Rolle’s theorem are satisfied on [x1 , x2 ]. Hence there must exist at least
one value x∗ ∈ (x1 , x2 ) such that f 0 (x∗ ) = 0. It therefore follows that betwen any two zeros x1
and x2 of f (x)there lies a ero x∗ of f 0 (x). Result 2: A polynomial of degree n can not have
more than n zeros.
Example Verify Rolle’s theorem for the function f (x) = x3 − 6x2 + 11x − 6
Solution
In this example we shall have to identify the closed interval [a,b] in which Rolle’s theorem is to
be verified. To determine a and b, we have to solve the equation f (x) = 0. That is,

x3 − 6x2 + 11x − 6 = 0
(x − 1)(x2 − 5x + 6) = 0
(x − 1)(x − 2)(x − 3) = 0

Therefore we obtain x = 1, 2, 3 ⇒ f (1) = f (2) = f (3) = 0. Choose a = 1 and b = 3 so

that the required interval is [1,3]. The function being a polynomial satisfies the requirements of
Rolle’s theorem for continuity and differentiability on the closed interval [1,3]. Further, since
f (1) = f (3), all conditions of Rolle’s theorem ae now satisfied. Therefore, the existence of

48
 atleast one c in the open interval (1,3) such that f 0 (c) = 0 is guaranteed. To determine the
point c, solve for x in the equation f 0 (x) = 0. √ √
That is f 0 (x) = 3x2 − 12x + 11 = 0 ⇒ x = 2 − 33 or x = 2 + 33 .
√ √
Thus we get two points c = 2 − 33 and c = 2 + 33 in the interval (1,3) such that f 0 (c) = 0.
Note In the above example we could have chosen intervals [1,2] or [2,3] instead of [1,3]. Rolle’s
theorem can be verified for these intervals also. it is for this reason that we get two values of c,
one lying in (1,2) and the other lying in (2,3).

5.0.7 Lagrange’s Mean Value Theorem

If a function f (x) is continuous on the closed interval [a,b] and differentiable on the open
interval (a,b), then there exists at least one point in (a,b) such that

f (b) − f (a)
(5.0.1) f 0 (c) =
b−a

(a)
(c)
(b)
3. Proof. The entire proof is divided into two parts;
2.
1.
Step I
Construct a new function F (x) with the help of the the given function which satisfies all the
requirements of Rolle’s theorem on the closed interval [a,b].
Step II
Prove Result I by using Rolle’s theorem on F (x) and closed interval [a,b].
We now prove the result,
Step I: Construction of F (x)
Assume that f (a) 6= f (b). For if f (a) = f (b) then the theorem reduces to Rolle’s theorem that
has already been discussed.
(graph) The curve APB represents the function f (x) on an interval [a,b]. The equation of the

chord AB through the points A(a, f (a)) and B(b, f (b) is given by

f (b) − f (b)
(5.0.2) y = f (a) + (x − a)
b−a
If P (x, f (x)) is an arbitrary point on the graph of the given function then P Q = P S − QS.
Since Q(x, y) is a point on the chord AB, then QS is given by

f (b) − f (b)
QS = f (a) + (x − a)
b−a
Therefore h f (b) − f (b) i
P Q = P S − QS = f (x) − f (a) + (x − a)
b−a
Since P Q varies as P varies, it follows that P Q is a function of x. Denote it by F (x) so that
h f (b) − f (b) i
(5.0.3) F (x) = f (x) − f (a) + (x − a) .
b−a
Equation (3) gives the required function.
Step II: Use of Rolle’s Theorem
From (3), it can be observed that the function F (x) can be considered to be a linear combination

49
of two functions f (x) − f (a) and x − a. Since f (x) − f (a) and x − a are both continuous on
the closed interval [a, b] and differentiable on the open interval (a, b) we have that
[(i)]F (x) is continuous on [a,b] F (x) is differentiable on (a,b)

Further, since
h f (b) − f (b) i
F (a) = f (a) − f (a) + (a − a) = 0,
b−a
h f (b) − f (b) i
F (b) = f (b) − f (a) + (a − a) = 0,
b−a
we have F (a) = F (b).
Thus all theree requirements for Rolle’s theorem are satisfied by the function F (x). Hence,
there must exist a value c in the interval (a,b) such that F 0 (c) = 0.
But
f (b) − f (a)
F 0 (x) = f 0 (x) − ,
b−a
f (b)−f (b) f (b)−f (b)
So F 0 (c) = 0 ⇒ f 0 (c) − b−a = 0 or f 0 (c) = b−a

Examples

[(i)]Find the value c of the Lagrange,s MVT when f (x) = x(x − 1) on the interval [0,2].
Solution
Since the given function is a polynomial in x, it is continuous for each x ∈ R. This
ensures the continuity of f (x) on [0,2]. Since a polynomial can be differentiable at each
x ∈ R, f 0 (x) exists at each x ∈ [0, 2]. Thus, there exists a value c in (0,2) such that

f (2) − f (0)
f 0 (c) = =1
2−0
But f 0 (x) = 2x − 1 ⇒ f 0 (c) = 2c − 1 = 1 ⇒ c = 1 Verify L.M.V.T for the function
f (x) = x(x − 1)(x − 2) on [0, 12 ].
Solution
The given function is a polynomial in x and can be written as f (x) = x3 − 3x2 + 2x.
Like in the previous example rhe given function is a polynomial and hence it satisfies the
conditions of the L.M.V.T on the interval [0, 12 ]. To determine the value of c we observe
that f (a) = f (0) = 0 and f (b) = f ( 12 ) = 38 . Also, f 0 (x) = 3x2 − 6x + 2. Now using the
L.M.V.T we have that
3
−0 3
3c2 − 6c + 2 = 81 =
2 − 0 4
√ √
⇒ c = 1+ 21
− 6 . The value c = 1 + 21
6 is not in the (0, 12 ) and therefore it is not required.
√
The required value is c = 1 − 621 .

5.0.8 Cauchy’s Mean Value Theorem (C.M.V.T)

let f (t) and g(t) be two functions continuous on the closed interval [a,b] and differentiable on
open interval (a,b). Let g 0 (t) 6= 0 for any t ∈ (a, b). Then there exists at least one value
c ∈ (a, b) such that
f (b) − f (a) f 0 (c)
= 0
g(b) − g(a) g (c)

50
2. Proof. The proof is divided into three steps
1.
[Step I]Construction of a new function F (t).
We construct a new function F (t) with the help of the given functions f (t) and
g(t) in such a way that it satisfies the conditions of Rolle’s theorem. Thus we
take
F (t) = [f (b) − f (a)][g(t) − g(a)] − [g(b) − g(a)][f (t) − f (a)].
We verify that F (t) satisfies the conditions of Rolle’s theorem.
Since f (t) and g(t) are continuous on [a,b] and differentiable on (a,b), F (t) being
linear combination of them will also be continuous on [a,b] and differentiable
on (a,b).
F (a) = [f (b) − f (a)][g(a) − g(a)] − [g(b) − g(a)][f (a) − f (a)] = 0 and F (b) = [f (b) −
f (a)][g(b) − g(a)] − [g(b) − g(a)][f (b) − f (a)] = 0. Therefore F (a) = f (b). Thus all
the conditions of Rolle’s theorem are satisfied by F (t) on [a,b].Apply Rolle’s
theorem to F (t) on [a,b].
There will exist a value c ∈ (a, b) such that F 0 (c) = 0. But F 0 (t) = [f (a) −
f (a)]g 0 (t) − [g(b) − g(a)]f 0 (t). Therefore, F 0 (c) = 0, ⇒ [f (a) − f (a)]g 0 (t) − [g(b) −
g(a)]f 0 (t) = 0. Hence,
f (b) − f (a) f 0 (c)
= 0
g(b) − g(a) g (c)

Remark: L.M.V.T is a special case of C.M.V.T, setting g(t) = t in Cauchy,s formula yields

f (b) − f (a)
f 0 (c) =
b−a

Examples

3. Discus C.M.V.T for the pair of functions g(x) = x2 and f (x) = x3 on the interval [−2, 3].
2.
1.
Solution
Clearly the functions g(x) = x2 and f (x) = x3 are continuous on [-2,3] and differentiable
on (-2,3). If the condition g 0 (x) 6= 0 for any x ∈ (−2, 3) is not necessary, then C.M.V.T
will be applicable to the pair g(x) = x2 and f (x) = x3 in the interval [-2,3] and therefore
there should exist a value x ∈ (−2, 3) such that

f 0 (x) f (3) − f (−2) 3x2 27 + 8 2

= ⇒ = =7 ⇒x=4 .
g 0 (x) g(3) − g(−2) 2x 9−4 3

But x = 4 32 is not an interior point to (-2,3). C.M.V.T has therefore failed. This failure
is essentially due to the fact that condition g 0 (x) 6= 0 is violated at the point x = 0 whichis
an interior point of (-2,3). This example clearly illustrate that the condition g 0 (t) 6= 0 for
any t ∈ (a, b) in the statement of C.M.V.T is a necessary condition.
2. For the pair cos x and sin x of functions, determine c using auchy’s formula on the interval
[0, π2 ].
Solution
Functions g(x) = cos x and f (x) = sin x are both continuous on [0, π2 ] and differentable
on (0, π2 ) hence the requirements of C.M.V.T are satisfied. Then there will exist a value
c such that 0 < c < π2 and

f 0 (c) f ( π2 − f (0) cos c 1−0 π

= ⇒ = ⇒ cot c = 1 ⇒ c = .
g 0 (c) g( π2 − g(0) − sin c 0−1 4

51
5.0.9 Indeterminate Forms

An expression is said to be in an indeterminate form if the limit of that expression is not evident
by inspection. Recall that
f (x) limx→a f (x)
lim =
x→a g(x) limx→a g(x)
Thus, if limx→a f (x) = limx→a g(x) = 0, then we say that the quotient function given by
f (x)/g(x) has the indeterminate form 0/0 at x = a.
Example
Consider limx→0 sinx x = 00 . Hence, sinx x has the indeterminate form 00 at x = 0.

L’Hopital’s Rule

Here we are interested in obtaining a method to evaluate the indeterminate form 00 . The fol-
lowing result known as L’Hopital’s rule is important in this connection.
Rule: Let c be a real number, +∞ or −∞ and f and g be two functions that satisfy
[(i)]f and g are differentiable at every point in the neighborhood of c, except possibly at c
0
itself (deleted neighborhood of c). limx→c f (x) = 0 = limx→c g(x) lim x→c fg0 (x)
(x)
= L, where
L is a real number, +∞ or −∞.
Then
f (x)
lim x→c = L.
g(x)

In otherwords under the hypothesis of the theorem the limit of the quotient of the two functions
is equal to the limit of the quotient of their derivatives. That is

f (x) f 0 (x)
lim x→c = lim x→c 0 .
g(x) g (x)

Examples

Evaluate the following limits

[(a)]
x2 − 1
lim ,
x→1 x − 1

Solution
We first note that limx→1 x2 − 1 = 0 = limx→1 x − 1. So that the given limit expression is
of the indeterminate form 00 , hence we can apply L’Hopital’s rule; that is
d 2
x2 − 1 dx (x − 1) 2x
lim = lim d
= lim =2
x→1 x − 1 x→1 1
dx (x − 1)
x→1

x − sin x
lim .
x→0 x2
Solution
The given limit expression is of the form 00 . So we apply L’Hopital’s rule. Therefore;

x − sin x 1 − cos x  0 
lim 2
= lim still form .
x→0 x x→0 2x 0

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 Apply L’Hopital’s rule again to get apply
1 − cos x sin x
lim = lim = 0.
x→0 2x x→0 2

Remarks; It must be carefully noted that

It many situations the limit expression remains of the form 00 even after the application
of L’Hopital’s rule. In such cases repeated application of the rule is made untill the
indeterminancy disapppears. The rule is generalised as follows. If in addition to the
three conditions in L’Hopital’s rule, we have f 0 (c) = f 00 (c) = ... = f n−1 (c) = 0 and
g 0 (c) = g 00 (c) = ... = g n−1 (c) = 0, then

f (x) f n (x)
lim x→c = lim x→c n
g(x) g (x)

provided the limit on the right exits.

+∞
−∞ form

+∞ 0
L’Hopital’s rule can be also applied if indeterminancy is of the form −∞ instead of 0
Example
Evaluate
log sin2 x
lim .
x→0 cot x
Solution
sin2 x
We notice that limx→0 logcot x
∞
= −∞ .
So, apply L’Hopital’s rule;

log sin2 x d
dx log sin2 x 2 cot x
lim = lim d
= lim =0
x→0 cot x x→0
dx cot x x→0 −cosec2 x

Other indeterminate forms

∞
L’Hopital’s rule is not only applied to 00 or +
− ∞ forms but at the same time the rule can usefuly
be employed for evaluating other indeterminate forms of the types ∞.0, 1∞ , 00 and ∞ − ∞.
The fundemental idea is that the given indeterminate form must be put either in the form 00 or
+∞
− ∞ by using some techniques before L’Hopital’s rule is applied.
Type ∞.0
If limx→c f (x) = 0 and limx→c g(x) = ∞ then limx→c f (x).g(x) is of the form 0.∞. Now

f (x)
lim [f (x).g(x)] = lim 1
x→c x→c
g(x)

and is now of the form 00 . Or

g(x)
lim [f (x).g(x)] = lim 1
x→c x→c
f (x)

and is of the form ∞∞

Type 1∞
If limx→c f (x) = 1 and limx→c g(x) = ∞ then limx→c f (x)g(x) is of the form 1∞ .
 

Let
y = lim f (x)g(x) .
 
x→c

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Then
log y = log lim f (x)g(x) = lim log f (x)g(x) = lim log g(x). log f (x).
   
x→c x→c x→c

This is of type ∞.0. The limit of the R.H.S of the above expression can be evaluated as follows.
Let it be L such that
log y = L ⇒ y = eL
.
The types 00 and ∞0 are similarly evaluated.
Type ∞ − ∞
If limx→c f (x) = ∞ = limx→c g(x), then limx→c [f (x) − g(x)] is of the form ∞ − ∞. To find y,
we have
 f (x)−g(x)   1 − 1 
f (x).g(x) g(x) f (x)
y = lim [f (x) − g(x)] = lim 1 = lim 1 .
x→c x→c x→c
f (x).g(x) f (x).g(x)
0
This is of type 0. Therefore, the above limit expression can be evaluated by application of
L’Hopital’s rule.

Examples

Evaluate the following limits

1 1 2
[(a)]limx→0 x log x limx→0 cos x) x2 limx→1 (1 − x2 ) log(1−x) limx→0 xx limx→0 (x + ex ) x
limx→0 (sec x − tan x)

Soluyion
(a) limx→0 x log x is of type ∞.0. Therefore
d 1
log x dx log x x
lim x log x = lim 1 = lim = lim = lim −x = 0
x→0 x→0 x→0 d 1 x→0 − 12 x→0
x dx x x

1 1
(b)limx→0 cos x) x2 is of type 1∞ . Let y = limx→0 cos x) x2 , then
d
1 log(cos x) dx log(cos x)
log y = lim 2
log(cos x) = lim = lim d 2
x→0 x x→0 x2 x→0
dx x

d d
dx log(cos x) − tan x dx − tan x − sec2 x 1
lim d 2
= lim = lim d
= lim =−
x→0
dx x
x→0 2x x→0
dx 2x
x→0 2 2
1
⇒ log y = − 12 or y = e− 2 . Therefore,
1 1
lim cos x) x2 = e− 2
x→0

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 2 2
(e)limx→0 (x + ex ) x is of type ∞0 . Therefore, we let y = limx→0 (x + ex ) x

2 log(x + ex )
log y = lim
x→0 x
d
log(x + ex )
=2 lim dx d
dx x
x→0

(1 + ex )
=2 lim
x→0 x + ex
(ex )
=2 lim
x→0 1 + ex
(ex )
=2 lim
x→0 1 + ex
(ex )
=2 lim x = 2
x→0 e

Therefore
2
lim (x + ex ) x = e2 .
x→0

5.0.10 Definition

Given the functin f (x) and its first n derivatives at x0 , Taylor’s polynomial for f (x) around x0
is given by
n
X f (k) (x0 )
Pn (x) = (x − x0 )k
k!
k=0

By construction f (x) and Pn (x) agree at x0 but if x is different from x0 say x∗ , then f (x∗ ) and
Pn (x∗ ) will mot be equal however small the difference f (x∗ ) − Pn (x∗ ) may be. Therefore, in
calculating the value of the function at x∗ using Taylor’s polnomial, some error will be introduced
into the final result. It is therefore very important to have some estimate for the error commited
in calculating the value of the function at the point close to x0 from the Taylor’s polynomial
Pn (x).

Taylor’s Theorem

Let f (x) be a function continuous on a closed interval [a,b]. If f (x0 ), f 0 (x0 ), f 00 (x0 ), ..., f n (x0 ),
x0 ∈ [a, b] are all given, then for any x ∈ [a, b], there exists a real number c between x0 and x
such that
n
X f (k) (x0 )
f (x) = (x − x0 )k + A(x − x0 )n+1 ,
k!
k=0

f (n+1) (c)
where A = − x0 )(n+1) .
(n+1)! (x
Remark
From Taylor’s theorem Rn (x) is given by

f (n+1) (c)
Rn (x) = (x − x0 )(n+1) .
(n + 1)!

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 Theorem

Under the assumption of Taylor’s theorem, there exists a real number M such that

(x − x0 )(n+1)
|Rn (x)| ≤ M for every x ∈ [a, b]
(n + 1)!

6. Proof. It is given that the (n + 1)t h derivative f (n+1) (x) is continuous on the closed interval
5.
4.
3.
2.
1.
[a,b], therefore, f (n+1) (x) is bounded below and above on [a,b]. Let M be the upper bound of
f (n+1) (x) on [a,b] , then

f (n+1) (c) f (n+1) (c)

|Rn (x)| = | (x − x0 )(n+1) | because x > x0 = | |(x − x0 )(n+1) .
(n + 1)! (n + 1)!
Thus,
(x − x0 )(n+1)
|Rn (x)| ≤ M .
(n + 1)!
This inequality provides an upper bound of the error and therefore we can only say the error
M
cannot exceed the value (n+1)! (x − x0 )n+1 .

Example

Find the Taylor polynomial of degree n about x0 for the following functions and find the maxi-
mum error introduced.
p
[(a)]f (x) = sin x, n = 4, x0 = π2 f (x) = (1 + x), n = 5, x0 = 0

Solution
(a)
n
X f (k) (x0 )
Pn (x) = (x − x0 )k
k!
k=0
π
Here f (xx) = sin x, n = 4, x0 = 2
⇒ f (x0 ) = sin π2 = 1, f 0 (x) = cos x ⇒ f 0 (x0 ) = cos π2 = 0, f 00 (x) = − sin x ⇒ f 00 (x0 ) =
− sin π2 = −1 f 000 (x) = − cos x ⇒ f 000 (x0 ) = − cos π2 = 0 f (iv) (x) = sin x ⇒ f (iv) (x) = sin π2 =
1
Thus,
4
X f (k) ( π2 )
P4 (x) = (x − π2)k so that
k!
k=0

sin x ≈P4 (x)

pi pi pi pi pi
=f ( ) + f 0 ( )(x − π2) + f 00 ( )(x − π2)2 + f 000 ( )(x − π2)3 + f (iv) ( )(x − π2)4
2 2 2 2 2
1 pi 1 pi
=1 − ( )(x − π2)2 + ( )(x − π2)4
2 2 4! 2
Hence
1 π 1 π
sin x = 1 − ( )(x − π2)2 + ( )(x − π2)4 + R5 (x),
2 2 4! 2
where
f (5) (c)  π 5 cos c  π 5
R5 (x) = x− = x−
5! 2 5! 2

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