Nationale Higher Analysis I Année Universitaire

School of Mathematics Chapter 4 2023-2024

1st Year Real Functions: Limits

0.1 Generalities
A real function f : R → R assigns for the independent variable x ∈ R at most one value y ∈ R. Usually
 write y = f (x). The set Df = {x ∈ R : f (x)2 is defined}
we  is called the domain of f and the set f (Df ) =
f (x) : x ∈ Df is the range of f. The subset of R , Γ (f ) = (x, f (x)) , x ∈ Df is the graph of the function f.
This graph may be sketched in the xy-coordinate plane, using y =  f (x). In other words, with each pair (x, f (x))
x
is associated a point M in xy-coordinate plane having coordinates f (x) .

Definition 1 Let f : Df → R be a function where its domain Df is symmetric about 0. f is said to be even if for
all x ∈ Df , f (−x) = f (x), and it is said to be odd if for all x ∈ Df , f (−x) = −f (x).
Geometric interpretation :
If f is even, then its graph Γ (f ) is symmetric about y-axis, and if f is odd, then Γ (f ) is symmetric about the
origin O.
In other words, if M is the point in the plane having its coordinates (x, f (x)) with x ∈ Df and M 0 is the point
in the plane having its coordinates (−x, f (−x)) , then f is even leads to M 0 is the symmetric of M relatively to
the y−axis, and f is odd leads to M 0 is the symmetric of M relatively to the origin of the plane O.
Examples 2

• f (x) = xn with n ∈ N, is even if n is even and is odd if n is odd.

• f (x) = cos x is even and f (x) = sin x is odd.
Definition 3 Let f : Df → R be a function. We say that f is est periodic if there exists a real number p > 0 such
that for all x ∈ Df , k ∈ Z, we have x + kp ∈ Df and f (x + kp) = f (x).

If the real number p0 = inf p > 0, ∀x ∈ Df f (x + p) = f (x) exists and is positive, then p0 is called the
period of f.
Examples 4

• f (x) = cos x is periodic with period equals 2π.

• f (x) = x − E(x) is periodic with period equals 1.
• The function f given by 
1, if x ∈ Q
f (x) = 0, if x ∈ R r Q
is periodic but it hasn’t a period.

Definition 5 Let f : Df → R be a function. we say that f is

• upper bounded if there exists M ∈ R such that for all x ∈ Df , f (x) ≤ M,

• lower bounded if there exists m ∈ R such that for all x ∈ Df , f (x) ≥ m,
• bounded if it is upper and lower bounded.
Definition 6 Let f : Df → R be a function and let I ⊂ Df be an interval in R. We say that f is

• nondecreasing on I if for all x, y ∈ I, x ≤ y implies f (x) ≤ f (y),

• nonincreasing on I if for all x, y ∈ I, x ≤ y implies f (x) ≥ f (y),
• monotonic on I if it is nondecreasing or nonincreasing on I.
• increasing on I if for all x, y ∈ I, x < y implies f (x) < f (y),
• decreasing on I if for all x, y ∈ I, x < y implies f (x) > f (y),
• strictly monotonic on I if it is increasing or decreasing on I.

Definition 7 Let λ be a real number, I be an interval of R and let f and g be two functions defined on I. The
f
functions λf, f + g, f · g, and are defined for any x ∈ I by
g
(λf ) (x) = λf (x),
(f + g) (x) = f (x) + g(x),
(f · g) (x) = f (x) · g(x),
f f (x)
(x) = , whenever g(x) 6= 0.
g g(x)

Definition 8 Let f : I → R and g : J → R be two functions such that f (I) ⊂ J. The composition of f by g is
the function denoted by g ◦ f and defined on I by g ◦ f (x) = g(f (x)).

Definition 9 Let I and J be two subsets of R with J ⊂ I and let f : I → R be a function defined on I. The
function fJ : J → R defined on J by fJ (x) = f (x) for all x ∈ J, is called the restriction of f to the subset J and
the function f is called an extension of fJ to the set I.

Definition 10 Let f : I → R be a function defined on the interval I. f is said to be a piece function if there
is n functions fi : Ii → R where the intervals I1 , ..., In satisfy ∪i=n
i=1 Ii = I and Ik ∩ Ij = ∅ if k 6= l such that
f (x) = fi (x) if x ∈ Ii .

Examples 11 The function f : R → R defined by

x2 , if x < 0,
(
f (x) = x, if x ∈ [0, 1] ,
1, if x > 1,
is a piecewise defined function.

0.2 Limits

0.2.1 Definitions

Definition 12 Let f : R → R be a function and let a be a real number. The function f is said to be defined

• in a neighborhood of a if there exists h > 0 such that f is defined on [a − h, a[ ∪ ]a, a + h] ,

• at right of a if there exists h > 0 such that f is defined on ]a, a + h] ,
• at left of a if there exists h > 0 such that f is defined on [a − h, a[ ,
• in a neighborhood of +∞ if there exists A > 0 such that f is defined on [A, +∞[ ,
• in a neighborhood of −∞ if there exists A > 0 such that f is defined on ]−∞, −A] .

Definition 13 Let f : R → R be a function and let a be a real number. We say that :

 1. f has l ∈ R as a limit when x tends to a if f is defined in neighborhood of a and for all  > 0 there exists
δ > 0 such that for all x ∈ R r {a} , |x − a| < δ implies |f (x) − l| < . In this case we write lim f (x) = l.
x→a

2. f has +∞ as a limit when x tends to a if f is defined in neighborhood of a and for all A > 0 there exists
δ > 0 such that for all x ∈ R r {a} , |x − a| < δ implies f (x) > A. In this case we write lim f (x) = +∞.
x→a

3. f has −∞ as a limit when x tends to a if f is defined in neighborhood of a and for all A > 0 there exists
δ > 0 such that for all x ∈ Rr{a} , |x − a| < δ implies f (x) < −A. In this case we write lim f (x) = −∞.
x→a

4. f has l ∈ R as a limit when x tends to a by great values if f is defined at right of a and for all  > 0 there
exists δ > 0 such that for all x ∈ R, 0 < x−a < δ implies |f (x) − l| < . In this case we write lim f (x) = l
>
x→a
or lim+ f (x) = l.
x→a

5. f has l ∈ R as a limit when x tends to a by least value if f is defined at left of a and for all  > 0 there exists
δ > 0 such that for all x ∈ R, 0 < a − x < δ implies |f (x) − l| < . In this case we write lim f (x) = l ou
<
x→a
lim f (x) = l.
x→a−

6. f has +∞ as a limit when x tends to a by great values if f is defined at right of a and for all A > 0 there
exists δ > 0 such that for all x ∈ R, 0 < x − a < δ implies f (x) > A. In this case we write lim f (x) = +∞
>
x→a
or lim+ f (x) = +∞.
x→a

7. f has +∞ as a limit when x tends to a by least values if f is defined at left of a and for all A > 0 there exists
δ > 0 such that for all x ∈ R, 0 < a − x < δ implies f (x) > A. In this case we write lim f (x) = +∞ or
<
x→a
lim f (x) = +∞.
x→a−

8. f has −∞ as a limit when x tends to a by great values if f is defined at right of a and for all A > 0
there exists δ > 0 such that for all x ∈ R, 0 < x − a < δ implies f (x) < −A. In this case we write
lim f (x) = −∞ or lim+ f (x) = −∞.
> x→a
x→a

9. f has −∞ as a limit when x tends to a by least values if f is defined at left of a and for all A > 0 there exists
δ > 0 such that for all x ∈ R, 0 < a − x < δ implies f (x) < −A. In this case we write lim f (x) = −∞ or
<
x→a
lim f (x) = −∞.
x→a−

Definition 14 Let f : R → R be function and a be a real number. We say that f has

• +∞ as a limit when x tends to +∞ if f is defined in a neighborhood of +∞ and for all A > 0 there exists
B > 0 such that for all x ∈ R, x > B implies f (x) > A. In this case we write lim f (x) = +∞.
x→+∞

• l ∈ R as a limit when x tends to +∞ if f is defined in a neighborhood of +∞ and for all  > 0 there exists
B > 0 such that for all x ∈ R, x > B implies |f (x) − l| < . In this case we write lim f (x) = l.
x→+∞

• −∞ as a limit when x tends to +∞ if f is defined in a neighborhood of +∞ and for all A > 0 there exists
B > 0 such that for all x ∈ R, x > B implies f (x) < −A. In this case we write lim f (x) = −∞.
x→+∞

• +∞ as a limit when x tends to −∞ if f is defined in a neighborhood of −∞ and for all A > 0 there exists
B > 0 such that for all x ∈ R, x < −B implies f (x) > A. In this case we write lim f (x) = +∞.
x→−∞

• l ∈ R as a limit when x tends to −∞ if f is defined in a neighborhood of −∞ and for all  > 0 there exists
B > 0 such that for all x ∈ R, x < −B implies |f (x) − l| < . In this case we write lim f (x) = l.
x→−∞

• −∞ as a limit when x tends to −∞ if f is defined in a neighborhood of −∞ and for all A > 0 there exists
B > 0 such that for all x ∈ R, x < −B implies f (x) < −A. In this case we write lim f (x) = −∞.
x→−∞
Theorem 15 The limit is unique.
Proof. Let f : R → R be a function and without loss of generality suppose that f has two finite limits l1 and
l2 when x tends to a. For  > 0, there is a neighborhood V of a, such that for all x ∈ R, x ∈ V , implies
|f (x) − l1 | <  and |f (x) − l2 | < . Hence, for any x ∈ V , we have
|l1 − l2 | ≤ |f (x) − l1 | + |f (x) − l2 | ≤ 2.
Since  is arbitrary, we conclude that l1 = l2 .
Remark 16 Let f : R → R be a function defined in a neighborhood of a ∈ R. Notice that lim f (x) = l if and
x→a
only if lim+ f (x) = lim− f (x) = l.
x→a x→a

Theorem 17 Let f : R → R a function defined in a neighborhood of a. Then lim f (x) = l if and only if for any
x→a
sequence (xn ), lim xn = a implies lim f (xn ) = l.
Proof. Suppose that lim f (x) = l and let (xn ) be a sequence such that lim xn = a. Without loss of generality
x→a
suppose that a, l ∈ R. Let  > 0, there exists δ > 0 such that for all x ∈ R, 0 < |x − a| ≤ δ implies
|f (x) − l| ≤ . Since lim xn = a, there exists nδ ∈ N such that |xn − a| ≤ δ for all n ≥ nδ . This leads to
|f (xn ) − l| ≤  for all n ≥ nδ . Hence we have proved that lim f (xn ) = l.
Suppose now that for any sequence (xn ) with lim xn = a, we have lim f (xn ) = l and let us prove that
lim f (x) = l. By the contrary, suppose lim f (x) 6= l. There exists 0 > 0 such that for all n ∈ N there is
x→a x→a
un such that |un − a| ≤ n1 and |f (un ) − l| > 0 . Thus we checked a sequence (un ) such that lim un = a and
lim f (un ) 6= l. This contradicts the hypothesis in the theorem.

Remark 18 Theorem 17 holds if a is replaced by a+ or a− .

0.2.2 Algebraic Proprieties of the limit

Proposition 19 Let f : R → R be a function defined in a neighborhood of a. Then, lim f (x) = l ∈ R if and only
x→a
if lim |f (x) − l| = 0.
x→a

Proof. For any sequence (xn ) satisfying lim xn = a, Proposition 2.9 of chapter real sequences, claim that
lim f (xn ) = l if and only if lim |f (xn ) − l| = 0. The proposition is proved.
Proposition 20 Let f and g be two functions defined in a neighborhood V of a such that f (x) ≤ g (x) for all
x ∈ V. If lim f (x) and lim g (x) exist then lim f (x) ≤ lim g (x) .
x→a x→a x→a x→a

Proof. For any sequence (xn ) ⊂ V such that lim xn = a, we have f (xn ) ≤ g (xn ) , Proposition 2.10 of chapter
real sequences, claim that lim f (xn ) = lim f (x) ≤ lim g (xn ) = lim g (x) . The proposition is proved.
x→a x→a

Proposition 21 Let f, g and h be three functions defined in a neighborhood V of a such that g (x) ≤ f (x) ≤
h (x) for all x ∈ V. Iflim g(x) and lim h (x) exist and lim g(x) = lim h (x) , then lim f (x) exists and lim f (x) =
x→a x→a x→a x→a x→a x→a
lim g (x) = lim h (x) .
x→a x→a

Proof. For any sequence (xn ) ⊂ V with lim xn = a, we have f (xn ) ≤ g (xn ) . Proposition 2.11 of chapter real
sequences, claim that
lim f (xn ) = lim f (x) ≤ lim g (xn ) = lim g (x) .
x→a x→a
Ending the proof.
1 sin x 1
Example : We have −1 ≤ sin x ≤ 1 for all x ∈ R. So we obtain − ≤ ≤ if x > 0. Since
x x x
sin x
lim − x1 = lim x1 = 0 we get lim


= 0.
x→+∞ x→+∞ x→+∞ x
Proposition 22 Let f : R → R be a function defined in a neighborhood of a. If lim f (x) = l, then lim |f (x)| =
x→a x→a
|l| . The reciprocal is true only if l = 0.
Proof. Indeed, 0 ≤ ||f (x)| − |l|| ≤ |f (x) − l| and the result becomes from the above proposition. If l = 0, we
have ||f (x)| − |l|| = ||f (x)|| = |f (x)| . The reciprocal is false because if one considers the function f (x) =
(−1)E(x) , lim |f (x)| = 1 but lim f (x) does not exist (lim f (x) = −1 and lim f (x) = 1).
x→1 x→1 >
x→1
<
x→1

Proposition 23 ( Algebraic operations ) Let λ ∈ R and let f, g : R → R be two functions defined in a neighbo-
rhood of a such that lim f (x) = l and lim g(x) exist. We have then
x→a x→a

 λl, if l ∈ R,

0, if λ = 0 and l = ±∞,
1) lim (λf (x)) = +∞, if l = +∞ and λ > 0 or l = −∞ and λ < 0,
x→a 
−∞, ( if l = −∞ et λ > 0 or l = +∞ et λ < 0.

l + k, if l, k ∈ R,
2) lim (f (x) + g (x)) = +∞, if l 6= −∞ et k = +∞ or l = +∞ et k 6= −∞,
x→a
 −∞, if l 6= +∞ et k = −∞ or l = −∞ et k 6= +∞..
 l · k, if l, k ∈ R,
 l > 0 and k = +∞ or l = +∞ and k > 0
+∞, if or l < 0 and k = −∞ or l = −∞ and k < 0,

3) lim (f (x) · g (x)) =
x→a
 −∞, if l > 0 and k = −∞ or l = −∞ and k > 0


or l < 0 and k = +∞ or l = +∞ and k < 0.
l


 , if l ∈ R and k ∈ R∗ ,

 k
f (x)  l > 0 and k = 0+ or l < 0 if k = 0− or
4) lim = +∞ if l = +∞ and k > 0 or l = −∞ if k < 0,
x→a g (x) 
+ −
 −∞, if l < 0 and k = 0 or l > 0 if k = 0



l = −∞ and k > 0 or l = +∞ if k < 0.

Proof. The proposition is obtained directly from Theorem 17 and Proposition 2.16 of chapter real sequences.
0 ∞
Remark 24 The cases ∞ − ∞, 0 · ∞, et are called undefined forms. The fact that they are avoid by the
0 ∞
above proposition does not mean that we can not compute them. There is useful technics for such situations, see
the following exercise.
Exercice 1 Compute the following limits.
√ √
x2 + 1 x−1 √ 
lim 3 , lim , lim x+1− x .
x→+∞ x + x + 2 x→1 x − 1 x→+∞

Solutions :    
2 1 1
x 1+ 2 1+ 2
x2 + 1 x 1 x
1. lim 3 = lim   = lim   = 0 · 1 = 0.
x→+∞ x + x + 2 x→+∞ 1 2 x→+∞ x 1 2
x3 1 + 2 + 3 1+ 2 + 3
x x x x
√ √
√

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

√ √
x+1− x x+1+ x 1
3. lim x + 1 − x = lim √ √ = lim √ √ = 0.
x→+∞ x→+∞ x+1+ x x→+∞ x+1+ x
Some standard limits :
You have already seen the following limits.
sin x ln x ln (1 + x)
lim = 1, lim = 0 et lim = 1.
x→0 x x→+∞ x x→0 x
Exercice 2 Compute the following limits.
1
1 − cos2 x x sin x − cos x + 1
lim , lim , lim (1 + x) x .
x→0 x2 x→0 x2 x→0

Solution :
2
1 − cos2 x sin2 x

sin x
1. lim = lim 2 = lim = 1.
x→0 x2 x→0 x x→0 x
   
x sin x − cos x + 1 x sin x 1 − cos x sin x (1 − cos x) (1 + cos x)
2. lim = lim + = lim +
x→0 x2 x→0 x2 x2 x→0 x x2 (1 + cos x)
sin x 1 − cos2 x 1 − cos2 x
 
1 sin x 1 1 3
= lim + 2
= lim + lim 2
lim =1+1· = .
x→0 x x 1 + cos x x→0 x x→0 x x→0 1 + cos x 2 2
1    
ln(1+x) ln(1+x)
3. lim (1 + x) x = lim exp x
= exp lim x = exp(1) = e.
x→0 x→0 x→0

0.2.3 Limits of monotonic functions

Theorem 25 Let f : (a, b) → R be a monotonic function, then the limits lim f (x) and limf (x) exist.
> <
x→a x→b

Proof. Suppose that f is nondecreasing (in the case f is nonincreasing we consider the function−f ). Let F =
{f (x), a < x < b} . We distinguish two cases.
1st case. F is upper bounded and set l = sup F. For any  > 0 there is x ∈ ]a, b[ such that f (x ) > l − .
Since f is nondecreasing we conclude that for all x ∈ ]x , b[
l −  < f (x ) ≤ f (x) < l + ,
that is limf (x) = l.
<
x→b

2nd case. F is not upper bounded. In this case for all A > 0 there exists xA ∈ ]a, b[ such that f (xA ) > A. Since
f is nondecreasing we conclude that for all x ∈ ]xA , b[ , f (x) > A. This means that limf (x) = +∞.
<
x→b

ln x ln x 1 − ln x
Examples 26 Let us show that limx→+∞ = 0. Set f (x) = , we have f 0 (x) = < 0 for
x x x
x > e. The function being decreasing, the limx→0 f (x) exists and because f (x) > 0 for x > 1. Therefore,
limx→+∞ f (x) = l where l is areal number. To find l we make use of Theorem 17 and properties of the function
ln.
ln(10n ) n ln(10)
For un = 10n , we have f (un ) = n
= and because 10n = (1+9)n = 1n +9n+ n(n−1)2
92 +... ≥
10 10n
n ln(10) n ln(10) ln(10)
n(n − 1) it follows that f (un ) = n
≤ = → 0 when n → +∞.
10 n(n − 1) n−1
ln x
Hence we have proved that limx→+∞ = 0.
x

0.3 Complements

1. Impact of the sign of the limit.

Proposition 27 Let f : R → R be a function. If limx→a f (x) = l with l > 0, then there is a neighborhood V of a
such that f (x) > 0 for all x ∈ V.
Proof. Without loss of generality, suppose that a, l ∈ R. From the definition of the limit, there exists δ > 0 such
that |f (x) − l| ≤ 2l for all x ∈ [a − δ, a) ∪ (a, a + δ]. Hence, for all x ∈ [a − δ, a) ∪ (a, a + δ] , we have
l 3l
0< ≤ f (x) ≤ .
2 2

2. Cauchy criterion.
Theorem 28 (Cauchy criterion) Let f : (a, b) → R be a function. The following assertions are equivalents.
1. lim− f (x) = l ∈ R.
x→b

2. For all ε > 0 there exists δ > 0 such that for all real numbers x, x0 ∈ (a, b) , x, x0 > b − δ implies
|f (x) − f (x0 )| < ε.
Proof. Suppose that lim− f (x) = l ∈ R, then for any ε > 0, there exists δ1 > 0 such that for all x ∈ (a, b),
x→b
b − x < δ implies |f (x) − l| < 2ε . Hence, for any x, x0 ∈ (a, b), x, x0 > b − δ implies |f (x) − f (x0 )| ≤
|f (x) − l| + |f (x0 ) − l| < 2ε + 2ε = ε.
Now, let (xn ) be a sequence such that lim xn = b. Let ε > 0 and let δ > 0 the positive real number given by
Assertion 2. Since (xn ) is a Cauchy sequence, there is nδ ∈ N such that for all n, m ∈ N, n, m ≥ nδ implies
|xm − xn | < δ. By means of 2, we obtain that |f (xm ) − f (xn )| < ε for all m, n ≥ nδ . This shows that (f (xn )) is
a Cauchy sequence and so it converges.
At let (un ) and (vn ) be two sequences such that lim un = lim vn = b and lim f (un ) = l and lim f (vn ) =
k. Let ε > 0 let δ > 0 the positive real number given by Assertion 2. There is Nδ ∈ N such that for all
n ∈ N, n ≥ Nδ implies that max(b − un , b − vn ) < 2δ . Therefore, for all n ∈ N, n ≥ Nδ implies that
|un − vn | ≤ (b − un ) + (b − vn ) < δ. So Assertion 2 leads to |f (un ) − f (vn )| < ε for all n ≥ Nδ . Passing to the
limit, we get |k − l| < ε. Because ε is aarbitrary, we conclude that k = l. Thus we have proved that all sequences
(f (xn )) where with lim xn = b, have the same limit. This proves that limx→b− f (x) exists and is finite.
3. Sup and inf limits. Let f : R → R be a function defined in a neighborhood (may be at right or at left) of a
and set 

 (a − δ, a + δ) , if a ∈ R,
 (δ, +∞) , if a = +∞,

Ia,δ = (−∞, δ) , if a = −∞,


 (a, a + δ) , if a ∈ R and f is defined at right of a,
(a − δ, a) , if a ∈ R and f is defined at left of a.


S(f, a, δ) = sup f (x) : x ∈ Ia,δ if f is upper bounded and

I(f, a, δ) = inf f (x) : x ∈ Ia,δ if f is lower bounded.

Because that δ → S(f, a, δ) and δ → I(f, a, δ) are monotonic functions, the limits
limδ→0 S(f, a, δ), limδ→0 I(f, a, δ) if a ∈ R, limδ→+∞ S(f, a, δ), limδ→+∞ I(f, a, δ) if a = +∞ and
limδ→−∞ S(f, a, δ), limδ→−∞ I(f, a, δ) if a = −∞ exist.

Definition 29 The sup-limit and inf-limit of the function f when x tends to a, are the limit denoted by limx→a f (x)
and limx→a f (x) and defined by

limδ→0 S(f, a, δ), if f is upper bounded,
limx→a f (x) = +∞, if not,

limδ→0 I(f, a, δ), if f is lower bounded,
limx→a f (x) = +∞, if not,
if a ∈ R, 
limδ→+∞ S(f, a, δ), if f is upper bounded,
limx→+∞ f (x) = +∞, if not,

limδ→+∞ I(f, a, δ), if f is lower bounded,
limx→+∞ f (x) = −∞, if not,
if a = +∞ and 
limδ→−∞ S(f, a, δ), if f is upper bounded,
limx→−∞ f (x) = +∞, if not,

limδ→−∞ I(f, a, δ), if f is lower bounded,
limx→−∞ f (x) = −∞, if not.

The sup and inf limits have the following properties.

1. lim f (x) = l if and only if limx→a f (x) = limx→a f (x) = l.

2. limx→a (−f (x)) = −limx→a f (x) and limx→a (−f (x)) = −limx→a f (x).
3. lim f (x) = ±∞ if and only there is a sequence (xn ) such that lim xn = a and lim f (xn ) = ±∞.
4. lim f (x) = l ∈ R if and only there is a sequence (xn ) such that lim xn = a and lim f (xn ) = l, and for all
k > l, there is δ > 0 such that S(f, a, δ) < k.
5. f ≤ g implies limx→a f (x) ≤limx→a g(x) and limx→a f (x) ≤ limx→a g(x).
6. limx→a f (x)+limx→a g(x) ≤limx→a (f (x) + g(x)) and limx→a (f (x) + g(x)) ≤ limx→a f (x) + limx→a g(x).
7. If limx→a f (x) exists then limx→a f (x)+limx→a g(x) =limx→a (f (x) + g(x)) and limx→a (f (x) + g(x)) =
limx→a f (x) + limx→a g(x).
8. If limx→a f (x) exists then limx→a f (x)limx→a g(x) =limx→a (f (x)g(x)) and
limx→a (f (x)g(x)) = limx→a f (x)limx→a g(x).