MA1521 Calculus for Computing

Lecture 2

Wong Yan Loi

National University of Singapore

January 14, 2022

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 1 / 42
Calculus - study of rate of change
Differential Calculus and Integral Calculus
Isaac Newton (1642–1727)
Gottfried Wilhelm Leibniz (1646-1716) Weierstrass
Eudoxus of Cnidus (408-355 BC) - Method of Exhaustion
Archimedes (287 - 212 BC)
Wilhelm Weierstrass (1815 -1897) - Limit

Newton Leibniz Eudoxus Archimedes

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 2 / 42
Table of Contents

1 Limits

2 Continuity

3 Evaluation of limits

4 Limits at infinity

5 More on Limits

6 Squeeze (Sandwich) Theorem

7 Intermediate Value Theorem

8 The Precise Definition of the Limit of a Function

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 3 / 42
Chapter 1: Limits and Continuity
Read Thomas’ Calculus, Chapter 2.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 4 / 42
 Limits

Limits

Let f be a real-valued function defined on some interval I (e.g.

(a, b), or (a, b] or (a, ∞)). Let c be a point in I.
lim f (x) is the value that f (x) approaches when x
x→c −
approaches c from the left.
lim f (x) is the value that f (x) approaches when x
x→c +
approaches c from the right.
Let c be an interior point (i.e. not an end point). If
lim f (x) = lim+ f (x) = L ∈ R, we say that lim f (x) exist and
x→c − x→c x→c
has value L.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 5 / 42
 Limits

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 6 / 42
 Limits

Left Limit Right Limit Limit

c lim f (x) lim+ f (x) lim f (x) f (x)
x→c − x→c x→c
0
2
4
-3
6
Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 7 / 42
 Limits

Left Limit Right Limit Limit

c lim f (x) lim f (x) lim f (x) f (x)
x→c − x→c + x→c
0 0 0 0 0
2 2 2 2 0
4 3 DNE DNE 3
−3 DNE 3 DNE undefined
6 0 DNE DNE 0

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 8 / 42
 Continuity

Let f be a real-valued function defined on some interval I (e.g.

(a, b), or (a, b] or (a, ∞)). Let c be a point in I.
Continuity at a point
Case 1 c is an interior point
• f is continuous at x = c if
(i) lim f (x) exists,
x→c
(ii) lim f (x) = f (c).
x→c

Case 2 c is the left end-point

• f is continuous at x = c if
(i) lim+ f (x) exists,
x→c
(ii) lim+ f (x) = f (c).
x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 9 / 42
 Continuity

Case 3 c is the right end-point

• f is continuous at x = c if
(i) lim f (x) exists,
x→c −
(ii) lim f (x) = f (c).
x→c −

Continuity on an interval
• f is continuous on an interval if f is continuous at x = c for all
points c in I.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 10 / 42
 Continuity

Results
If f and g are continuous at x = c, then for any constant k and any
positive constant n, each of the following functions is continuous
at x = c.
(i) f + g, (ii) f n , (iii) kf , (iv) fg, (v) f /g provided g(c) 6= 0.

If g is continuous at x = c and f is continuous at x = g(c), then

the composite function f ◦ g is continuous at x = c.
(Note (f ◦ g)(x) = f (g(x))).

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 11 / 42
 Continuity

Example
Find the points of discontinuity of the function f whose graph on (−3, 6]
is given below.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 12 / 42
 Continuity

Solution.

Point of discontinuity Reason

x =2 lim f (x) 6= f (2)

x→2

x =4 lim f (x) does not exist

x→4

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 13 / 42
 Evaluation of limits

Results (Law of limits)

The following results are true provided all the limits involved exist. The
limit could be one-sided or two-sided. The number k is a constant.
1. lim (f (x) ± g(x)) = lim f (x) ± lim g(x)
x→c x→c x→c
2. lim kf (x) = k lim f (x)
x→c x→c
3. lim (f (x)g(x)) = ( lim f (x))( lim g(x))
x→c x→c x→c

f (x) lim f (x)

x→c
4. lim =
x→c g(x) lim g(x)
x→c
5. If g is continuous at the point b and lim f (x) = b, then
x→c
lim g(f (x)) = g(b) = g( lim f (x))).
x→c x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 14 / 42
 Evaluation of limits

Result. The following functions are continuous on any interval

contained in their maximal domain.
1. Polynomials
2. Trigonometric Functions
3. Exponential Functions
4. Logarithmic Functions
5. A combination of any of the above on the domain it is defined.
For example, a rational function P(x)/Q(x) is continuous at all points x
where Q(x) 6= 0.
Since lim f (x) = f (c) when f is continuous at x = c, finding the limit at
x→c
x = c of any of the above functions is a matter of evaluating f at x = c.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 15 / 42
 Evaluation of limits

Example
x + ln(x + 3)
Evaluate lim √ .
x→−2 x +6

x+ln(x+3)
Solution. The function √
x+6
is continuous at x = −2. Thus

x + ln(x + 3) −2 + ln(−2 + 3)
lim √ = √ = −1.
x→−2 x +6 −2 + 6


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 16 / 42
 Evaluation of limits

Exercise
Evaluate lim tan3 (sin x).
x→0

Ans: 0.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 17 / 42
 Limits at infinity

Let f be defined on R.
lim f (x) is the value f (x) approaches as x tends to positive
x→∞
infinity.
lim f (x) is the value f (x) approaches as x tends to negative
x→−∞
infinity.

Graphically, if lim f (x) = c ∈ R or lim f (x) = c, then the line y =

x→∞ x→−∞
c is a horizontal asymptote of the graph of f (x).

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 18 / 42
 Limits at infinity

1 1
For any positive integer n, lim = 0 and lim = 0.
x→∞ xn x→−∞ xn
1
For example, lim 2 − = 2, and y = 2 is a horizontal asymptote of
x→∞ x
the graph of y = 2 − x1 .

1
Also lim 2 − = 2.
x→−∞ x
Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 19 / 42
 Limits at infinity

Example
 2
3 p
Evaluate lim + 4 − e−x .
x→∞ 2x

Solution.
2
√

3 p
lim + 4 − e−x = (0 + 4)2 = 4.
x→∞ 2x


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 20 / 42
 Limits at infinity

Exercise
 
4
Evaluate lim ln 3 − 2 sin .
x→−∞ x

Ans: ln 3.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 21 / 42
 More on Limits

Indeterminate forms.
f (x)
(a) A limit of the form lim where f (x) → 0 and g(x) → 0 as
g(x)
x→c
x → c is called an indeterminate form of the type 00 .
f (x)
(b) A limit of the form lim where f (x) → ∞ and g(x) → ∞
x→c g(x)
as x → c is called an indeterminate form of the type ∞ ∞.

Replacement rule. Let I be an open interval containing the point

x = c. Suppose f (x) = g(x) for all x ∈ I, except possibly at x = c.
Then lim f (x) = lim g(x).
x→c x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 22 / 42
 More on Limits

Example
x 2 − 7x + 6
Evaluate lim .
x→6 36 − x 2

Solution.

x 2 − 7x + 6 (x − 1)(x − 6) x −1 5
lim 2
= lim = lim =− .
x→6 36 − x x→6 −(6 + x)(x − 6) x→6 −(6 + x) 12


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 23 / 42
 More on Limits

Exercise
√ √
x + 12 − 6 − x
Evaluate lim .
x→−3 18 − 2x 2

1
Ans: 36 .

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 24 / 42
 More on Limits

P(x)
Result. Limits of the form lim , where P(x) and Q(x) are
x→±∞ Q(x)
polynomials in x.

leading term 
z}|{  0 if α < β
Ax α A

P(x) +··· 
B if α = β
lim = lim =
x→±∞ Q(x) x→±∞ Bx β +···   ∞
| {z ∞} if α > β.
or −
|{z} 
leading term depends on the question

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 25 / 42
 More on Limits

Example
√
(18x 2 + 5x − 1)(2 x − 1)3
Evaluate lim .
x→∞ (3x − 1)4

√ 7
(18x 2 + 5x − 1)(2 x − 1)3 144x 2 + · · ·
Solution. lim = lim = 0.
x→∞ (3x − 1)4 x→∞ 81x 4 + · · ·


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 26 / 42
 More on Limits

Exercise
(1 + 2x)3
Evaluate lim √ .
x→−∞ 16x 6 + 9x − 1

Ans: −2.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 27 / 42
 More on Limits

Useful results.
If lim g(x) = 0, then
x→c
sin(g(x)) g(x)
lim = lim = 1,
x→c g(x) x→c sin(g(x))
tan(g(x)) g(x)
lim = lim = 1.
x→c g(x) x→c tan(g(x))
In particular, when c = 0 and g(x) = x,
sin x x
lim = lim = 1,
x→0 x x→0 sin x
tan x x
lim = lim = 1.
x→0 x x→0 tan x

sin 3x ln x sin(e−x )
For example, lim = 1, lim = 1, lim = 1.
x→0 3x x→1 tan(ln x) x→∞ e−x

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 28 / 42
 More on Limits

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 29 / 42
 More on Limits

Example
x tan(2x) + sin2 x
Evaluate lim .
x→0 sin(3x 2 ) + x tan(2x)

Solution. 2
x tan(2x) + sin2 x 2x 2 tan(2x)
2x + x 2 sinx 2 x
lim = lim
x→0 sin(3x 2 ) + x tan(2x)
2
x→0 3x 2 sin(3x ) + 2x 2 tan(2x)
2 3x 2x
2
2 tan(2x)
2x + sinx 2 x 2+1 3
= lim 2
= = .
x→0 3 sin(3x ) + 2 tan(2x) 3+2 5
3x 2 2x


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 30 / 42
 More on Limits

Exercise
 √ 
Evaluate lim+ x 2 cot(2x) csc2 (3 x) .
x→0

1
Ans: 18 .

Exercise
√
tan( x − 2)
Evaluate lim .
x→4 sin(16 − x 2 )

1
Ans: − 32 .

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 31 / 42
 Squeeze (Sandwich) Theorem

Squeeze Theorem I. Suppose g(x) ≤ f (x) ≤ h(x) for all x in some

open interval containing a point c, except possibly at x = c. If
lim g(x) = lim h(x) = L, then lim f (x) = L.
x→c x→c x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 32 / 42
 Squeeze (Sandwich) Theorem

Example
It is given that 3 − x 2 ≤ f (x) ≤ 1 + 2ex for all x. Find lim f (x).
x→0

Solution. As lim 3 − x 2 = 3 and lim 1 + 2ex = 3, we have by

x→0 x→0
squeeze theorem I that lim f (x) = 3.
x→0


Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 33 / 42
 Squeeze (Sandwich) Theorem

Exercise
Use Squeeze theorem to show that lim |f (x)| = 0 ⇒ lim f (x) = 0.
x→c x→c

Remark. The converse of the above result, namely

lim f (x) = 0 ⇒ lim |f (x)| = 0 is true.
x→c x→c
Hence, we have

Result. lim f (x) = 0 ⇔ lim |f (x)| = 0.

x→c x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 34 / 42
 Squeeze (Sandwich) Theorem

Squeeze Theorem II. If lim g(x) = 0, then for any function h,

x→c

lim g(x) sin(h(x)) = 0 and lim g(x) cos(h(x)) = 0.

x→c x→c

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 35 / 42
 Squeeze (Sandwich) Theorem

Example
 
2
Evaluate lim x 3 cos( ) .
x→0 sin x

Solution. Since lim x 3 = 0, we have by squeeze theorem II,

 x→0

3 2
lim x cos( ) = 0.
x→0 sin x


Exercise
 
1 1
Evaluate lim sin(2 ln x) + 2x sin( ) .
x→∞ ln x x

Ans: 2.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 36 / 42
 Intermediate Value Theorem

Theorem. (Intermediate Value Theorem) If f is continuous on [a, b]

and k is a number between f (a) and f (b), then f (c) = k for some
c ∈ [a, b].

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 37 / 42
 Intermediate Value Theorem

Example
Show that the equation x 3 ex = 10 has a root between 1 and 1.5.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 38 / 42
 Intermediate Value Theorem

Solution. Let f (x) = x 3 ex . f is continuous on R.

We have f (1) = e = 2.718 and f (1.5) = 1.53 e1.5 = 15.126.

Thus f (1) < 10 < f (1.5).

By Intermediate Value Theorem, there is a root to f (x) = 10 between 1

and 1.5.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 39 / 42
 Intermediate Value Theorem

Exercise
Show that the equation 10 = x + 2 tan(2x) has a root between 3 and 4.

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 40 / 42
 The Precise Definition of the Limit of a Function

Let f (x) be defined on an open interval containing the point c, except

possibly at c itself. We say that the limit of f (x) as x approaches c is the
number L, and write
lim f (x) = L,
x→c

if, for every number  > 0, there exists a corresponding number δ > 0
such that for all x,

0 < |x − c| < δ ⇒ |f (x) − L| < .

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 41 / 42
 The Precise Definition of the Limit of a Function

Example
Prove from definition that lim 5x − 3 = 2.
x→1

Solution. Note that |(5x − 3) − 2| = 5|x − 1|. Given  > 0, we choose

δ = /5. Then for all x,

0 < |x − 1| < δ ⇒ 5|x − 1| < 5δ ⇒ |(5x − 3) − 2| < .

Exercise
x
Prove from definition that lim + 3 = 1.
x→−6 3

Wong Yan Loi (NUS) MA1521 Calculus for Computing Lecture 2 January 14, 2022 42 / 42