Chapter 2
Differentiation of Functions of
one Real Variable
2.1
Differentiation
Definition: Let A ⊆ R and let f : A → R. Let a ∈ A. Then f is said
to be differentiable at x = a if
f (a + h) − f (a)
exists finitely.
h→0
h
lim
If the limit exists finitely, then it is called the derivative of f at x =a or
df
.
differential co-efficient of f at x = a and is denoted by f 0 (a) or
dx x=a
i.e.
f 0 (a)
=
=
lim
f (a + h) − f (a)
.
h
lim
f (x) − f (a)
.
x−a
h→0
x→a
(∵ x = a + h).
f (a + h) − f (a)
= f 0 (a) exists finitely, then
h→0
h
f (a + h) − f (a)
∀ ε > 0, ∃ δ > 0 such that
− f 0 (a)
h
satisfying 0 < |h − 0| < δ.
If lim
< ε for all h
Note: A function f is differentiable at x = a means geometrically that a
tangent can be drawn to the curve y = f (x) at x = a. The gradient or
slope of this tangent is equal to f 0 (a).
36
f (a + h) − f (a)
h→0
h
exists finitely, then we say that f is differentiable at x = a from the left.
Definition: Let f : A → R be a function and a ∈ A. If lim−
This finite limit is called the left derivative of f at x = a and is denoted
by f−0 (a).
f (a + h) − f (a)
.
h
i.e. f−0 (a) = lim−
h→0
f (a + h) − f (a)
h→0
h
exists finitely, then we say that f is differentiable at x = a from the right.
Definition: Let f : A → R be a function and a ∈ A. If lim+
This finite limit is called the right derivative of f at x = a and is denoted
by f+0 (a).
f (a + h) − f (a)
.
h
i.e. f+0 (a) = lim+
h→0
Theorem 2.1. Let f : A → R, A ⊆ R and let a ∈ A. Then f is
differentiable at x = a iff f−0 (a) = f+0 (a).
Eg: Discuss differentiability of the each of the following functions
f : R → R at the origin.
1. f (x) = sin x.
2. f (x) = xn
∀ n ∈ N.
3. f (x) = |x|.
√
x.

3 + x if x ≥ 0,
5. f (x) =
3 − x if x < 0.
4. f (x) =
Solution:
1. f (x) = sin x.
sin h − 0
.
h→0
h→0
h
sin h
= lim
.
h→0
h
= 1 (∈ R).
=⇒ f is differentiable at x = 0 and f 0 (0) = 1.
lim
f (0 + h) − f (0)
h
=
lim
37
2. f (x) = xn
∀ n ∈ N.
f (0 + h) − f (0)
h
lim
h→0
hn − 0
.
h→0
h
= lim hn−1 .
h→0

1 if n = 1,
=
0 if n 6= 1.
=
lim

1 if n = 1,
=⇒ f is differentiable at x = 0 and f 0 (0) =
0 if n =
6 1.
3. f (x) = |x|.
lim−
h→0
f (0 + h) − f (0)
h
=
lim−
|h| − 0
.
h
lim−
−h
.
h
h→0
=
h→0
= −1.
=⇒ f−0 (0) = −1.
lim+
h→0
f (0 + h) − f (0)
h
=
lim+
|h| − 0
.
h
lim+
+h
.
h
h→0
=
h→0
=
=⇒
f+0 (0)
1.
= 1.
Thus f−0 (0) 6= f+0 (0).
=⇒ f is not differentiable at x = 0.
4. f (x) =
√
x.
f (0 + h) − f (0)
lim
h→0
h
√
h−0
= lim
.
h→0
h
√
h
= lim
.
h→0 h
1
lim √ .
h→0
h
= ∞.
=⇒ f is not differentiable at x = 0.
=
38

3 + x if x ≥ 0,
5. f (x) =
3 − x if x < 0.
lim−
h→0
f (0 + h) − f (0)
h
=
lim−
h→0
3−h−3
.
h
= −1.
=⇒ f−0 (0) = −1.
lim+
h→0
f (0 + h) − f (0)
h
=
h→0
=
=⇒
f+0 (0)
lim+
3+h−3
.
h
1.
= 1.
Thus f−0 (0) 6= f+0 (0).
=⇒ f is not differentiable at x = 0.
Theorem 2.2. If a function f : A → R, A ⊆ R, is differentiable at x = a
then it is continuous at x = a.
Proof. Suppose that f is differentiable at x = a.
f (x) − f (a)
exists finitely and equal to f 0 (a).
x→a
x−a
f (x) − f (a)
.
i.e. f 0 (a) = lim
x→a
x−a
Then lim
For all x ∈ A (x 6= a), we have,
f (x) − f (a)
f (x) − f (a) =
· (x − a).
(x − a)
=⇒
f (x) − f (a)
lim [f (x) − f (a)]
= lim
· (x − a).
x→a
x→a
(x − a)
f (x) − f (a)
· lim (x − a).
= lim
x→a
x→a
(x − a)
= f 0 (a) · 0.
=
0.
=⇒ lim f (x) − f (a)
=
0.
lim f (x)
=
f (a).
x→a
x→a
Hence f is continuous at x = a.
39
Note: The converse part of the above theorem need not be true.
Eg: f : R → R given by f (x) = |x|, is not differentiable at x = 0.
Let ε > 0 be arbitrary.
Choose δ = ε.
Then δ > 0.
For all x satisfying |x − a| < δ, |f (x) − f (0)| = | |x| − 0| = |x| < δ = ε.
i.e. ∃ δ > 0 such that |f (x) − f (0)| < ε for all x satisfying 0 < |x − 0| < δ.
Since ε > 0 is arbitrary,
∀ ε > 0, ∃ δ > 0 such that |f (x) − f (0)| < ε for all x satisfying
0 < |x − 0| < δ.
Hence f is continuous at x = 0.
Theorem 2.3. Let A ⊆ R, let a ∈ A and f : A → R and g : A → R
be two functions such that these are differentiable at x = a.
1. If k ∈ R, then the function kf is differentiable at x = a and
(kf )0 (a) = kf 0 (a);
2. The function (f ± g) is differentiable at x = a and
(f ± g)0 (a) = f 0 (a) ± g 0 (a);
3. The function (f · g) is differentiable at x = a and
(f · g)0 (a) = f 0 (a) g(a) + f (a) g 0 (a);
4. If g(x) 6= 0 ∀ x ∈ A, then the function
and
f
is differentiable at x = a
g
0
g(a) f 0 (a) − f (a) g 0 (a)
f
(a) =
.
g
[g(a)]2
Proof.
1. f : A → R is differentiable at x = a and k ∈ R.
Case(i): k = 0.
In this case, the result is trivial.
Case(ii): k 6= 0.
Let ε > 0 be arbitrary.
Since f is differentiable at x = a,
f (x) − f (a)
= f 0 (a) exists.
x→a
x−a
lim
40
=⇒ ∃ δ > 0 such that
f (x) − f (a)
ε
− f 0 (a) <
for all x
x−a
|k|
satisfying 0 < |x − a| < δ.
=⇒ ∃ δ > 0 such that
k f (x) − k f (a)
− k f 0 (a) < ε for all x
x−a
satisfying 0 < |x − a| < δ.
Hence by the definition,
k f (x) − k f (a)
= k f 0 (a).
x→a
x−a
lim
Thus k f is differentiable at x = a and (kf )0 (a) = k · f 0 (a).
2. Let ε > 0 be arbitrary.
Since f is differentiable at x = a,
f (a + h) − f (a)
ε
∃ δ1 > 0 such that
− f 0 (a) <
for all h satisfyh
2
ing 0 < |h| < δ1 .
Since g is differentiable x = a,
g(a + h) − g(a)
ε
∃ δ2 > 0 such that
− g 0 (a) <
for all h satisfying
h
2
0 < |h| < δ2 .
Choose δ = M in(δ1 , δ2 ).
ε
f (a + h) − f (a)
− f 0 (a) <
&
h
2
g(a + h) − g(a)
ε
− g 0 (a) <
for all h satisfying 0 < |h| < δ.
h
2
Then ∃ δ > 0 such that
For all h satisfying 0 < |h| < δ,
=
=
≤
<
=
(f + g)(a + h) − (f + g)(a)
− (f 0 (a) + g 0 (a))
h
f (a + h) + g(a + h) − f (a) − g(a)
− (f 0 (a) + g 0 (a)) .
h
f (a + h) − f (a)
g(a + h) − g(a)
− f 0 (a) +
− g 0 (a) .
h
h
f (a + h) − f (a)
g(a + h) − g(a)
− f 0 (a) +
− g 0 (a) .
h
h
+ .
2 2
.
i.e ∃ δ > 0 such that
(f + g)(a + h) − (f + g)(a)
− (f 0 (a) + g 0 (a)) < ε for all h satisfyh
ing 0 < |h| < δ.
41
Since ε > 0 is arbitrary,
∀ ε > 0, ∃ δ > 0 such that
(f + g)(a + h) − (f + g)(a)
− (f 0 (a) + g 0 (a))
h
fying 0 < |h| < δ.
< ε for all h satis-
Hence by the definition,
f + g is differentiable at x = a and (f + g)0 (a) = f 0 (a) + g 0 (a).
Similarly we can prove that,
f − g is differentiable at x = a and (f − g)0 (a) = f 0 (a) − g 0 (a).
3. Since f & g are differentiable at x = a,
g(a + h) − g(a)
f (a + h) − f (a)
f 0 (a) = lim
& g 0 (a) = lim
.
h→0
h→0
h
h
lim
h→0
(f · g)(a + h) − (f · g)(a)
h
f (a + h) · g(a + h) − f (a) · g(a)
= lim
.
h→0
h
=
=
=
lim
f (a + h) · g(a + h) − f (a) g(a + h) + f (a) g(a + h) − f (a) · g(a)
.
h
lim
[f (a + h) − f (a)] · g(a + h) + f (a) · [g(a + h) − g(a)]
.
h
h→0
h→0
f (a + h) − f (a)
g(a + h) − g(a)
· lim g(a + h) + lim f (a) · lim
.
h→0
h→0
h→0
h→0
h
h
lim
= f 0 (a) · lim g(a + h) + lim f (a) · g 0 (a).
h→0
h→0
Since g is differentiable at x = a, g is continuous at x = a.
i.e. lim g(x) = g(a).
x→a
i.e. lim g(a + h) = g(a).
h→0
Then the equation (4.1) implies that,
lim
h→0
(f · g)(a + h) − (f · g)(a)
= f 0 (a) · g(a) + f (a) · g 0 (a).
h
Hence f · g is differentiable at x = a and
(f · g)0 (a) = f 0 (a) · g(a) + f (a).g 0 (a).
42
(2.1)
f
f
(a + h) −
(a)
g
g
4. lim
h→0
h
=
=
=
f (a + h) f (a)
−
g(a + h)
g(a)
lim
.
h→0
h
f (a + h)g(a) − f (a)g(a + h)
.
h→0
g(a) · g(a + h) · h
lim
f (a + h)g(a) − f (a)g(a + h) − f (a)g(a) + f (a)g(a)
.
h→0
g(a) · g(a + h) · h
lim
=
g(a)[f (a + h) − f (a)] − f (a)[g(a + h) − g(a)]
.
h→0
g(a) · g(a + h) · h
=
1
f (a + h) − f (a)
f (a)
g(a + h) − g(a)
lim
−
.
h→0 g(a + h)
h
g(a)g(a + h)
h
lim
f (a)
1
· f 0 (a) − lim
· g 0 (a).
h→0 g(a)g(a + h)
h→0 g(a + h)
= lim
Since g is differentiable at x = a, g is continuous at x = a.
1
1
=⇒ lim
=
.
x→a g(x)
g(a)
1
1
=
.
h→0 g(a + h)
g(a)
=⇒ lim
From (4.2), we have,
f
f
(a + h) −
(a)
g
g
lim
h→0
h
=
1
f (a)
· f 0 (a) −
· g 0 (a).
g(a)
[g(a)]2
=
g(a)f 0 (a) − f (a)g 0 (a)
.
[g(a)]2
f
Hence
is differentiable at x = a and
g
0
f
g(a)f 0 (a) − f (a)g 0 (a)
(a) =
.
g
[g(a)]2
43
(2.2)
Corollary: If f1 , f2 , · · · , fn are functions from A → R and A ⊆ R
and if these functions are differentiable at ‘a’, a ∈ A then,
1. the function f1 + f2 + · · · + fn is differentiable at ‘a’ and
(f1 + f2 + · · · + fn )0 (a) = f10 (a) + f20 (a) + · · · + fn0 (a).
2. the function f1 f2 · · · fn is differentiable at ‘a’ and
(f1 f2 · · · fn )0 (a) = f10 (a)f2 (a) · · · fn (a) + f1 (a)f20 (a) · · · fn (a) + · · ·
+ f1 (a)f2 (a) · · · fn0 (a).
Theorem 2.4. (Chain Rule): Let A, B ⊆ R. Let f : A → R and
g : B → R be two functions such that f (A) ⊆ B and let a ∈ A. If f is
differentiable at ‘a’ and g is differentiable at ‘f (a)’ then the composition
function g◦f is differentiable at ‘a’ and
(g◦f )0 (a) = g 0 (f (a)) · f 0 (a).
Proof. Suppose that f is differentiable at ‘a’ and g is differentiable at
f (a) (∈ B).
f (x) − f (a)
g(f (x)) − g(f (a))
Then lim
= f 0 (a) and
lim
= g 0 (f (a)).
x→a
f
(x)→f
(a)
x−a
f (x) − f (a)
Consider,
g◦f (x) − g◦f (a)
lim
x→a
x−a
g(f (x)) − g(f (a))
.
= lim
x→a
x−a
g(f (x)) − g(f (a)) f (x) − f (a)
·
.
x→a
f (x) − f (a)
(x − a)
=
lim
g(f (x)) − g(f (a))
f (x) − f (a)
· lim
.
x→a
x→a
f (x) − f (a)
x−a
= lim
(2.3)
Since f is differentiable at x = a, f is continuous at x = a.
Then
f (x) → f (a) as x → a.
(2.4)
From (4.3) & (4.4), we have,
g◦f (x) − g◦f (a)
=
x→a
x−a
lim
g(f (x)) − g(f (a))
f (x) − f (a)
· lim
.
x→a
f (x)→f (a)
f (x) − f (a)
(x − a)
lim
= g 0 (f (a)) · f 0 (a) .
Thus g◦f is differentiable at x = a and
(g◦f )0 (a) = g 0 (f (a)) · f 0 (a).
44
Definition:
A function f : A → R is said to be differentiable on A, if
f differentiable at every point in A.
Note:
If f : A → R is differentiable on A and g : B → R is dif-
ferentiable on f (A) then g◦f is differentiable on A.
2.2
Properties of Differentiable Functions
Definition: A function f : R → R is said to be strictly increasing at
x = a, if ∃ δ > 0 such that f (x) < f (a) for all x ∈ (a − δ , a) and
f (x) > f (a) for all x ∈ (a , a + δ).
Definition: A function f : R → R is said to be increasing at x = a, if
∃ δ > 0 such that f (x) 6 f (a) for all x ∈ (a − δ , a) and f (x) > f (a)
for all x ∈ (a , a + δ).
Eg: A function f : R → R is defined by f (x) = x ∀ x ∈ R.
f is strictly increasing function at every point which is greater than zero.
Definition: A function f : R → R is said to be strictly decreasing at
x = a, if ∃ δ > 0 such that f (x) > f (a) for all x ∈ (a − δ , a) and
f (x) < f (a) for all x ∈ (a , a + δ).
Definition: A function f : R → R is said to be decreasing at x = a, if
∃ δ > 0 such that f (x) > f (a) for all x ∈ (a − δ , a) and f (x) 6 f (a)
for all x ∈ (a , a + δ).
Eg: A function f : R → R is defined by f (x) = −x ∀ x ∈ R .
f is strictly decreasing function at every point which is less than zero.
Theorem 2.5. If f is differentiable at x = a and f 0 (a) > 0 then f is
strictly increasing at x = a.
Proof. Suppose that f is differentiable at x = a and f 0 (a) > 0.
f (x) − f (a)
Then lim
= f 0 (a) > 0.
x→a
x−a
f (x) − f (a)
=⇒ ∀ ε > 0, ∃ δ > 0 such that
− f 0 (a) < ε for all
x−a
x ∈ (a − δ , a) ∪ (a , a + δ).
45
Choose ε =
f 0 (a)
(> 0).
2
f (x) − f (a)
f 0 (a)
− f 0 (a) <
for all
x−a
2
x ∈ (a − δ , a) ∪ (a , a + δ).
Then ∃ δ > 0 such that
f 0 (a)
f (x) − f (a)
f 0 (a)
<
− f 0 (a) <
for all
2
x−a
2
x ∈ (a − δ , a) ∪ (a , a + δ).
=⇒ −
=⇒
f 0 (a)
f (x) − f (a)
f 0 (a)
<
<3
for all x ∈ (a − δ , a) ∪ (a , a + δ).
2
x−a
2
f (x) − f (a)
> 0 for all x ∈ (a − δ , a) ∪ (a , a + δ).
x−a
f (x) − f (a)
For all x ∈ (a − δ , a),
> 0 and for all x ∈ (a , a + δ),
x−a
f (x) − f (a)
>0.
x−a
=⇒ For all x ∈ (a − δ , a), f (x) − f (a) < 0 and
=⇒
for all x ∈ (a, a + δ), f (x) − f (a) > 0 .
i.e. ∃ δ > 0 such that f (x) < f (a) for all x ∈ (a − δ , a) and
f (x) > f (a) for all x ∈ (a, a + δ) .
Hence by the definition,
f is strictly increasing at x = a.
Note: Converse part of the above theorem need not be true.
i.e. f is strictly increasing at x = a =⇒ f 0 (a) > 0.
Eg: A function f : R → R defined by, f (x) = x3
∀ x ∈ R, is strictly
increasing at x = 0.
f 0 (0) = 0.
f is strictly increasing at x = 0 but f 0 (0) ≯ 0.
Note:
1. If f is differentiable at x = a and f 0 (a) < 0 then f is strictly
decreasing at x = a.
2. Converse part of the above result need not be true.
46
Extremum of Real-Valued Function
Definition: Let f be real-valued function defined on R and let a ∈ R.
Then the function f is said to have a local maximum(relative maximum) at
x = a, if ∃ δ > 0 such that f (x) < f (a) for all x ∈ (a−δ , a)∪(a , a+δ).
Similarly we say that the function f has a local minimum(relative minimum) at x = a, if ∃ δ > 0 such that f (x) > f (a) for all
x ∈ (a − δ , a) ∪ (a , a + δ).
Note:We say that f has a local extremum or relative extremum at x = a,
if it has either local maximum or local minimum at x = a.
Theorem 2.6. If a function f : R → R has a local extremum at
x=a
and
f is differentiable at x = a then f 0 (a) = 0.
Proof. Case(i): Suppose that f has a local maximum at x = a and f
is differentiable at x = a.
Since f is differentiable at x = a,
f 0 (a) exists finitely.
Now, there are 3 possibilities:
1. f 0 (a) > 0
2. f 0 (a) < 0
3. f 0 (a) = 0.
Suppose that f 0 (a) > 0.
By the last theorem, f is strictly increasing at x = a.
i.e. ∃ δ > 0 such that f (x) < f (a) ∀ x ∈ (a − δ , a) and f (x) > f (a)
∀ x ∈ (a , a + δ).
This contradicts to the fact that f has local maximum at x = a.
Hence f 0 (a) ≯ 0.
Suppose that f 0 (a) < 0.
By the last theorem, f is strictly decreasing at x = a.
i.e. ∃ δ > 0 such that f (x) > f (a) ∀ x ∈ (a − δ , a) and f (x) < f (a)
∀ x ∈ (a , a + δ).
This contradicts to the fact that f has local maximum at x = a.
Thus f 0 (a) ≮ 0.
Hence f 0 (a) = 0.
47
Case(ii): Suppose that f has a local minimum at x = a and f is
differentiable at x = a.
Since f is differentiable at x = a, f 0 (a) exists finitely.
Now, there are three possibilities,
1. f 0 (a) > 0
2. f 0 (a) < 0
3. f 0 (a) = 0.
Suppose that f 0 (a) > 0.
By the last theorem, f is strictly increasing at x = a.
i.e. ∃ δ > 0 such that f (x) < f (a) ∀ x ∈ (a − δ , a) & f (x) > f (a)
∀ x ∈ (a , a + δ).
This contradicts to the fact that f has local minimum at x = a.
Hence f 0 (a) ≯ 0.
Suppose that f 0 (a) < 0.
By the last theorem, f is strictly decreasing at x = a.
i.e. ∃ δ > 0 such that f (x) > f (a) ∀ x ∈ (a − δ , a) & f (x) < f (a)
∀ x ∈ (a , a + δ).
This contradicts to the fact that f has local minimum at x = a.
Thus f 0 (a) ≮ 0.
Hence f 0 (a) = 0.
In both cases f 0 (a) = 0.
Note:
1. Converse part of the above theorem need not be true.
i.e. f 0 (a) = 0 =⇒ f has a local extremum at x = a.
Eg: Consider the function f : R → R defined by f (x) = x3
∀ x ∈ R.
f 0 (x) = 3x2 .
f 0 (0) = 0.
If x > 0, f (x) > f (0).
If x < 0, f (x) < f (0).
f has no extremum at x = 0.
2. In the above theorem the statement that the function f is differentiable at x = a is essential.
48
Eg: Consider the function f : R → R defined by f (x) = |x| ∀ x ∈ R.
f has a local minimum at x = 0 and f is not differentiable at x = 0.
f 0 (0) does not exist.
3. A point x = a
at which f 0 (x) = 0 where f has no local extremum
at x = a , then x = a is called a point of inflexion.
Theorem 2.7. ( Rolle’s Theorem): Let A ⊆ R and f : A → R be a
function with [a , b] ⊆ A. Suppose that f is continuous on [a , b] and
differentiable on (a , b). If f (a) = f (b), then ∃ c ∈ (a , b) such that
f 0 (c) = 0.
Proof. Suppose that f is continuous on [a , b]
and differentiable on
(a , b).
Suppose also that f (a) = f (b).
Since f is continuous on [a , b], f is bounded on [a , b].
Then f attains its supremum & its infimum.
Let M = sup f (x) and m = inf f (x).
[a , b]
[a , b]
Then ∃ c , d ∈ [a , b] such that f (c) = M and f (d) = m.
Case(i): Both c and d are end points of [a , b].
f (c) = M, f (d) = m and f (a) = f (b).
=⇒ M = m.
=⇒ f (x) = M = m is constant function on [a , b].
=⇒ f 0 (x) = 0 ∀ x ∈ [a , b].
Hence the result follows.
Case(ii): At least one of c , d is not an end point of [a , b].
Without loss of generality we assume that c ∈ (a , b).
Since f is differentiable on (a , b), f 0 (c) exists.
1. Suppose that f 0 (c) > 0.
Then f is strictly increasing at x = c.
=⇒ ∃ δ > 0 such that f (x) < f (c) ∀ x ∈ (c − δ , c) and
f (x) > f (c) ∀ x ∈ (c , c + δ).
=⇒ f (x) < M ∀ x ∈ (c − δ , c) and f (x) > M ∀ x ∈ (c , c + δ).
Then the second inequality contradicts to the fact that M
supremum of f (x).
Hence f 0 (c) ≯ 0.
49
is the
2. Suppose that f 0 (c) < 0.
Then f is strictly decreasing at x = c.
=⇒ ∃ δ > 0 such that f (x) > f (c) ∀ x ∈ (c − δ , c) and
f (x) < f (c) ∀ x ∈ (c , c + δ).
=⇒ f (x) > M ∀ x ∈ (c − δ , c) and f (x) < M ∀ x ∈ (c , c + δ).
Then the first inequality contradicts to the fact that M is the supremum of f (x).
Hence f 0 (c) ≮ 0.
Therefore f 0 (c) = 0.
i.e. ∃ c ∈ (a , b) such that f 0 (c) = 0.
Eg: If f : R → R is differentiable at all points of R, then between any two
roots of f (x) = 0 there will be at least one root for f 0 (x) = 0.
Solution: Let x1 & x2 be any two roots of f (x) = 0.
Without loss of generality, we assume that x1 < x2 .
Since f is differentiable on R, f is continuous on R.
Then
(i) f is continuous on [x1 , x2 ].
(ii) f is differentiable on (x1 , x2 ).
(iii) f (x1 ) = f (x2 ) = 0.
By the Rolle’s theorem,
∃ c ∈ (x1 , x2 ) such that f 0 (c) = 0.
i.e. c is a root of f 0 (x) = 0.
Theorem 2.8. ( Mean Value Theorem): Let f : [a , b] → R be
a function. Suppose that f
is continuous on [a , b] and that f
is
differentiable on (a , b) . Then ∃ c ∈ (a , b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Proof. Suppose that f is continuous on [a , b] and that f is differentiable
on (a , b).
Define the function
Φ : [a , b] → R by
f (b) − f (a)
Φ(x) = f (x) −
(x − a) ∀ x ∈ [a , b].
b−a
50
Since f is continuous on [a , b] and (x − a) is continuous on [a , b],
Φ is continuous on [a , b].
Since f is differentiable on (a , b) and (x − a) is differentiable on (a , b),
Φ is differentiable on (a , b).
Further Φ(a) =
f (a),
f (b) − f (a)
Φ(b) = f (b) −
(b − a) = f (a).
b−a
Therefore Φ(a) = Φ(b).
By the Rolle’s theorem,
0
∃ c ∈ (a , b) such that
Φ (c) = 0.
f (b) − f (a)
0
0
i.e. Φ (c) = f (c) −
= 0.
b−a
i.e. ∃ c ∈ (a , b) such that f 0 (c) =
f (b) − f (a)
.
(b − a)
Corollary:
(i) If f : R → R is continuous on [a , b] and differentiable on (a, b)
and f 0 (x) = 0
∀ x ∈ [a , b], then f is a constant function on
[a , b].
Proof. Suppose that f is continuous on [a , b] and differentiable on
(a , b).
Let x1 , x2 be two points in [a , b].
Without loss of generality, we assume that x1 < x2 .
Since f is continuous on [a , b], f is continuous on [x1 , x2 ].
Since f is differentiable on (a , b), f is differentiable on (x1 , x2 ) .
By the Mean Value Theorem,
f (x2 ) − f (x1 )
.
x2 − x1
Since f 0 (x) = 0 ∀ x ∈ [a , b], f 0 (c) = 0.
f (x2 ) − f (x1 )
=⇒ f 0 (c) =
= 0.
x2 − x1
=⇒ f (x2 ) = f (x1 ).
∃ c ∈ (x1 , x2 ) such that f 0 (c) =
Since x1 & x2 are arbitrary points in [a , b],
f is a constant on [a , b].
51
(ii) Let f : [a , b] → R and g : [a , b] → R be two continuous functions
on [a , b]. Suppose that f & g are differentiable on (a , b) and
that f 0 (x) = g 0 (x) for all x ∈ [a , b], then ∃ a constant k such
that f (x) = g(x) + k
∀ x ∈ [a , b].
Proof. Consider the function f − g : [a , b] → R.
Since f and g are continuous on [a , b], (f − g) is continuous on
[a , b].
Since f and g are differentiable on [a , b], (f − g) is differentiable
on [a , b].
Since f 0 (x) = g 0 (x) ∀ x ∈ [a , b],
f 0 (x) − g 0 (x) = 0 ∀ x ∈ [a , b].
=⇒ (f 0 − g 0 )(x) = 0 ∀ x ∈ [a , b].
=⇒ (f − g)0 (x) = 0 ∀ x ∈ [a , b].
By corollary (i),
(f − g) is a constant function on [a , b] .
i.e. ∃ a constant k such that (f − g)(x) = k
∀ x ∈ [a , b].
=⇒ ∃ a constant k such that f (x) − g(x) = k
∀ x ∈ [a , b].
=⇒ ∃ a constant k such that f (x) = g(x) + k
∀ x ∈ [a , b].
Theorem 2.9. Let f : R → R be differentiable function on R such that
f 0 (c) = 0 for some c ∈ R. Suppose that f 00 (c) exists. Then,
1. if f 00 (c) > 0 then f has a local minimum at x = c.
2. if f 00 (c) < 0 then f has a local maximum at x = c.
Proof. Suppose that f is differentiable on R and f 0 (c) = 0 for some
c ∈ R.
1. Let f 00 (c) > 0 .
Then f 0 is strictly increasing at x = c.
=⇒ ∃ δ > 0 such that f 0 (c) > f 0 (x) ∀ x ∈ (c − δ , c) and
f 0 (c) < f 0 (x) ∀ x ∈ (c , c + δ).
i.e.
∃ δ > 0 such that f 0 (x) < 0 ∀ x ∈ (c−δ , c) and f 0 (x) > 0 ∀ x ∈ (c , c+δ).
(2.5)
52
Let x ∈ (c − δ , c) be an arbitrary element.
Since f is differentiable on [x , c], f is continuous on [x , c].
By the Mean Value Theorem,
∃ c0 ∈ (x , c) such that f 0 (c0 ) =
From (4.5) & (4.6), we have,
f (c) − f (x)
.
c−x
(2.6)
f (c) − f (x)
< 0.
c−x
=⇒ f (c) − f (x) < 0.
i.e. f (c) < f (x).
Since x is arbitrary,
f (c) < f (x)
∀ x ∈ (c − δ, c).
(2.7)
Let x ∈ (c , c + δ) be an arbitrary element.
Since f is differentiable on R, f is differentiable on (c , x) and f
is continuous on [c , x].
By the Mean Value Theorem,
∃ c00 ∈ (c , x) such that f 0 (c00 ) =
f (x) − f (c)
.
x−c
From (4.5),we have, f 0 (c00 ) > 0.
f (x) − f (c)
> 0.
x−c
=⇒ f (x) − f (c) > 0.
=⇒
i.e. f (x) > f (c).
Since x is arbitrary,
f (c) < f (x)
∀ x ∈ (c , c + δ).
(2.8)
From (4.7) & (4.8), we obtain that,
∃ δ > 0 such that f (c) < f (x) ∀ x ∈ (c − δ , c) ∪ (c , c + δ).
Hence by the definition,
f has a local minimum at x = c.
2. Let f 00 (c) < 0.
Then f 0 is strictly decreasing at x = c.
=⇒ ∃ δ > 0 such that f 0 (x) > f 0 (c) ∀ x ∈ (c − δ , c) and
f 0 (x) < f 0 (c) ∀ x ∈ (c , c + δ).
53
(2.9)
i.e. ∃ δ > 0 such that f 0 (x) > 0 ∀ x ∈ (c − δ , c) and f 0 (x) < 0
∀ x ∈ (c , c + δ).
Let x ∈ (c − δ , c) be any arbitrary element.
Since f is differentiable on R, f is differentiable on (x , c).
By the Mean Value Theorem,
∃ c1 ∈ (x , c) such that f 0 (c1 ) =
f (c) − f (x)
.
c−x
From (4.9),we have, f 0 (c1 ) > 0.
f (c) − f (x)
>0.
c−x
=⇒ f (c) > f (x).
=⇒
Since x is arbitrary,
f (c) > f (x) ∀ x ∈ (c − δ , c).
(2.10)
Let x ∈ (c , c + δ) be any element.
Since f is differentiable on R, f is differentiable on (c , x).
By the Mean Value Theorem,
∃ c2 ∈ (c , x) such that f 0 (c2 ) =
f (x) − f (c)
.
x−c
From (4.9),we have, f 0 (c2 ) < 0.
f (x) − f (c)
<0.
x−c
=⇒ f (x) < f (c).
=⇒
Since x is arbitrary,
f (x) < f (c) ∀ x ∈ (c , c + δ).
From (4.10) & (4.11), we obtain that,
∃ δ > 0 such that f (x) < f (c) ∀ x ∈ (c − δ , c) ∪ (c , c + δ).
Hence by the definition,
f has a local maximum at x = c.
54
(2.11)
Eg:
1. If f : R → R is a twice differentiable on R such that f (a) = f (b) = 0
and f (c) > 0 for some c ∈ (a , b), then ∃ at least a ρ ∈ (a , b)
such that f 00 (ρ) < 0.
Solution: Since f is differentiable on R, f is differentiable on (a , c)
and f is continuous on [a , c].
By the Mean Value Theorem,
∃ x ∈ (a , c) such that f 0 (x) =
f (c) − f (a)
f (c)
=
> 0.
c−a
c−a
i.e. f 0 (x) > 0.
Again f is differentiable on [c , b], f is continuous on [c , b].
By the Mean Value Theorem,
∃ y ∈ (c , b) such that f 0 (y) =
f (b) − f (c)
−f (c)
=
< 0.
b−c
b−c
i.e. f 0 (y) < 0.
Since f is twice differentiable function on R, f 0 is differentiable on
R.
=⇒ f 0 is differentiable on (x , y) and f 0 is continuous on [x , y].
By the Mean Value Theorem,
f 0 (y) − f 0 (x)
< 0.
∃ ρ ∈ (x , y) such that f (ρ) =
y−x
00
i.e. ∃ ρ ∈ (a , b) such that f 00 (ρ) < 0.
2. If R → R is a twice differentiable on R such that f (a) = f (b) for
some a , b ∈ R. If a and b are two consecutive roots of the equation
f 00 (x) = 0 then, ∃ a unique element ρ ∈ (a , b) such that f has a
maximum or minimum at x = ρ.
Solution: Since a and b are two consecutive roots of the equation
f 00 (x) = 0,
f 00 (a) = 0 & f 00 (b) = 0 and f 00 (x) 6= 0 ∀ x ∈ (a , b).
Since f is differentiable on R, f is differentiable on (a , b) and f
is continuous on [a , b].
By the Mean Value Theorem,
∃ ρ ∈ (a , b) such that f 0 (ρ) =
f (b) − f (a)
= 0.
b−a
55
(∵ f (a) = f (b) )
Since f 00 (x) 6= 0
∀ x ∈ (a , b),
f 00 (ρ) 6= 0.
We have f 0 (ρ) = 0 and f 00 (ρ) 6= 0.
i.e. f 0 (ρ) = 0 and f 00 (ρ) > 0 or f 00 (ρ) < 0.
Hence f has a maximum or minimum at x = ρ.
Suppose that ρ1 , ρ2 ∈ (a , b) such that f 0 (ρ1 ) = f 0 (ρ2 ) = 0.
Assume that ρ1 6= ρ2 .
Without loss of generality we assume that ρ1 < ρ2 .
Since f 0 is differentiable on R, f 0 is differentiable on (ρ1 , ρ2 ) and
f 0 is continuous on [ρ1 , ρ2 ].
By the Mean Value Theorem,
∃ ρ ∈ (ρ1 , ρ2 ) such that f 00 (ρ) =
f 0 (ρ2 ) − f 0 (ρ1 )
= 0.
ρ2 − ρ1
i.e. ∃ ρ ∈ (a , b) such that f 00 (ρ) = 0, a contradiction.
=⇒ ρ1 = ρ2 .
Thus, ∃ a unique ρ ∈ (a , b) such that f has a maximum or
minimum at x = ρ.
Note: The Mean Value Theorem is sometimes given in the following convenient form:
f (a + h) = f (a) + f 0 (c)h, where c ∈ (a , a + h)
or
f (a + h) = f (a) + h f 0 (a + θh), where θ ∈ (0 , 1).
x
Eg: Show that x < sin−1 x < √
∀ x ∈ (0 , 1).
1 − x2
Let f : [0 , 1] → R be defined by f (x) = sin−1 x ∀ x ∈ [0 , 1].
Then f is continuous on [0 , 1] and differentiable on (0 , 1).
Then f is continuous on [0 , x] and differentiable on (0 , x),
x ∈ (0 , 1).
Then by the Mean Value Theorem,
f (x) − f (0)
∃ c ∈ (0 , x) such that f 0 (c) =
.
x−0
f (x)
=⇒ f 0 (c) =
.
x
1
f (x)
i.e. √
=
.
2
x
1−c
56
where
Since 0 < c < x,
1
1
1< √
<√
.
1 − c2
1 − x2
f (x)
1
i.e. 1 <
<√
.
x
1 − x2
x
=⇒ x < sin−1 x < √
1 − x2
∀ x ∈ (0 , 1).
Theorem 2.10. (Cauchy Mean Value Theorem): Suppose that both
functions f and g are continuous on [a , b] and differentiable on (a , b).
Suppose also that g 0 (x) 6= 0 ∀ x ∈ (a , b). Then
f (b) − f (a)
f 0 (c)
= 0 .
g(b) − g(a)
g (c)
Proof. Note that since g 0 (x) 6= 0 for x in (a , b),
it follows from the Rolle’s Theorem that g(a) 6= g(b).
Define Φ : [a , b] → R by f (b) − f (a)
Φ(x) = f (x) −
(g(x) − g(a)) ∀ x ∈ [a , b].
g(b) − g(a)
Since f & g are continuous on [a , b], Φ is continuous on [a , b].
Since f & g are differentiable on (a , b), Φ is differentiable on (a , b).
Further, we have,
Φ(a)
= f (a),
Φ(b) = f (a).
Thus Φ(a) = Φ(b).
Then by the Rolle’s Theorem, ∃ c ∈ (a , b) such that Φ0 (c) = 0.
f (b) − f (a) 0
i.e. ∃ c ∈ (a , b) such that f 0 (c) −
g (c) = 0.
g(b) − g(a)
f (b) − f (a)
f 0 (c)
i.e. ∃ c ∈ (a , b) such that
= 0
g(b) − g(a)
g (c).
Theorem 2.11. (Taylor’s Theorem):Suppose that f : R → R and its
derivatives f (1) , f (2) , · · · , f (n−1) are continuous on [a , a + h] and that
f (n) exists on (a , a + h).
h (1)
h2 (2)
Then f (a + h) = f (a) + f (a) +
f (a) + · · ·
1!
2!
n−1
h
hn (n)
+
f (n−1) (a) +
f (a + θh),
(n − 1)!
n!
where θ ∈ (0 , 1).
57
Proof. Define Φ : [0 , h] → R by
t
tn−1
tn
Φ(t) = f (a + t) − f (a) − f (1) (a) − · · · −
f (n−1) (a) − B;
1!
(n − 1)!
n!
where we choose B to make Φ(h) = 0.
From the definition of Φ, we see that
Φ(0) = 0, Φ(1) (0) = 0 = Φ(2) (0) = · · · = Φ(n−1) (0).
Since f, f (1) , f (2) , · · · , f (n−1) are continuous on [a , a + h],
Φ, Φ(1) , Φ(2) , · · · , Φ(n−1) are continuous on [a , a+h] and Φ, Φ(1) , Φ(2) , · · · , Φ(n−1)
are differentiable on (a , a + h).
Now, we will apply the Rolle’s Theorem to Φ, Φ(1) , Φ(2) , · · · , Φ(n−1) n times.
Since Φ(0) = 0 = Φ(h), by the Rolle’s Theorem,
∃ h1 ∈ (0 , h) such that Φ(1) (h1 ) = 0.
Since Φ(1) (0) = 0 = Φ(1) (h1 ), by the Rolle’s Theorem to Φ(1) ,
∃ h2 ∈ (0 , h1 ) such that Φ(2) (h2 ) = 0.
Continuing this way, we obtain that ,
∃ hn ∈ (0 , hn−1 ) such that Φ(n) (hn ) = 0.
Since 0 < hn < hn−1 < · · · < h2 < h1 < h,
hn = θh, where 0 < θ < 1.
Now Φ(n) (t) = f (n) (a + t) − B.
=⇒ Φ(n) (hn ) = f (n) (a + hn ) − B.
=⇒ B = f (n) (a + θh),
(∵ hn = θh).
Then we have,
t (1)
f (a) − · · ·
1!
tn−1
tn
−
f (n−1) (a) − f (n) (a + θh).
(n − 1)!
n!
Φ(t) = f (a + t) − f (a) −
(2.12)
By substituting t = h in (4.12), we have,
h (1)
h2 (2)
f (a + h) = f (a) + f (a) +
f (a) + · · ·
1!
2!
hn−1
hn (n)
+
f (n−1) (a) +
f (a + θh),
(n − 1)!
n!
(2.13)
where 0 < θ < 1 .
Note:
1. The term Rn (h) =
hn (n)
f (a + θh) is called the Lagrange form of ren!
mainder.
58
2. The equation (4.13) can be written as
(x − x0 ) (1)
(x − x0 )2 (2)
f (x) = f (x0 ) +
f (x0 ) +
f (x0 ) + · · ·
1!
2!
(x − x0 )n−1 (n−1)
(x − x0 )n (n)
+
f
(x0 ) +
f (c),
(n − 1)!
n!
where c ∈ (x0 , x).
Note: Taylor’s Theorem with Cauchy’s form of the remainder
If f : R → R has a continuous (n − 1)th derivative on [a , a + h] and is
differentiable n times on (a , b) then
h2 (2)
hn−1
h
f (a) + · · · +
f (n−1) (a)
f (a + h) = f (a) + f (1) (a) +
1!
2!
(n − 1)!
hn
(1 − θ)n−1 f (n) (a + θh),
+
(n − 1)!
where 0 < θ < 1.
Theorem 2.12. (Maclaurin’s Theorem): By putting a = 0 and h = x in
the Taylor’s Theorem, we obtain for f : [0 , x] → R, which is differentiable
n times, that
x (1)
x2 (2)
xn−1
xn (n)
f (0) +
f (0) + · · ·+
f (n−1) (0) +
f (θx),
1!
2!
(n − 1)!
n!
for some θ ∈ (0 , 1).
f (x) = f (0)+
1
Eg: 1 − x2 6 cos x for all x ∈ R.
2
Let f (x) := cos x.
Using the Taylor’s Theorem, with x0 = 0, we obtain,
1
cos x = 1 − x2 + R2 (x),
2
f (3) (c) 3
where R2 (x) =
x , 0 < c < x,
3!
sin c 3
x.
6
If 0 6 x 6 π, then 0 < c 6 π.
=
=⇒ R2 (x) > 0.
Also if − π 6 x 6 0, then − π 6 c < 0.
Then R2 (x) > 0.
1 2
x 6 cos x for |x| 6 π.
2
1
If |x| > π, then 1 − x2 < −3 6 cos x.
2
1 2
Hence 1 − x 6 cos x for all x ∈ R.
2
Therefore 1 −
59
Eg: For any k ∈ N, and for all x > 0, we have,
1
1 2k
1
1
x − x2 + · · · −
x < ln(1 + x) < x − x2 + · · · +
x2k+1 .
2
2k
2
2k + 1
Let f (x) = ln(1 + x), x > 0.
Then
f (1) (x)
f (2) (x)
f (3) (x)
..
.
f (n) (x)
1
.
1+x
1
= −
.
(1 + x)2
2
2!
=
=
.
3
(1 + x)
(1 + x)3
=
=
(−1)n−1
(n − 1)!
.
(1 + x)n
Then the Taylor’s Theorem with x0 = 0 is given by,
(x − x0 )2 (2)
(x − x0 )n (n)
(x − x0 ) (1)
f (x0 ) +
f (x0 ) + · · · +
f (x0 ) + Rn (x),
f (x) = f (x0 )+
1!
2!
n!
where Rn (x) =
(−1)n f (n+1) (c)
(x − x0 )n+1 , where c ∈ (0 , x).
(n + 1)!
=⇒ f (x) = x −
1 2
(−1)n−1 xn
x + ··· +
+ Rn (x),
2
n
where Rn (x) =
1
(−1)n
xn+1 .
n + 1 (1 + c)n+1
Thus for any x > 0, if n = 2k, then R2k (x) > 0 and if
n = 2k + 1, R2k+1 (x) < 0.
Hence the required inequality follows immediately.
2.3
L’ Hospital’s Rules.
Indeterminate Forms
It was shown before that if lim f (x) = ` and lim g(x) = m, and if m 6= 0,
x→a
x→a
f (x)
`
then lim
= .
x→a g(x)
m
If m = 0, ` 6= 0, then the limit is infinite.
f
0
For the case ` = m = 0, the limit of is , called indeterminate form.
g
0
We will see that in this case the limit may not exist or may be any real value,
depending on the functions f and g.
For example, if we define f (x) = α x, α ∈ R,
f (x)
αx
= lim
= α.
lim
x→0 x
x→0 g(x)
60
and g(x) = x, then
Note: Other indeterminate forms are represented by the symbols
∞
, 0 · ∞ , 00 , 1∞ , ∞0 , ∞ − ∞. Our attention will be focused on the
∞
0
∞
indeterminate forms
and
. The other indeterminate forms are usually
0
∞
0
∞
reduced to the form
and
.
0
∞
L’Hospital’s Rule: The case
0
.
0
Theorem 2.13. Suppose that f and g are continuous on [a , b], differentiable on (a , b) and that f (c) = g(c) = 0 for some c ∈ [a , b] and
that g(x) 6= 0 & g 0 (x) 6= 0 for all x ∈ [a , b] \ {c}.Then
f 0 (x)
f (x)
= L (∈ R), then lim
= L.
0
x→c g (x)
x→c g(x)
1. if lim
f 0 (x)
= +∞ (or − ∞), then
x→c g 0 (x)
2. if lim
f (x)
= +∞ (or − ∞).
x→c g(x)
lim
Proof. (1). Suppose that f and g are continuous on [a , b] and differentiable on (a , b) and that f (c) = g(c) = 0 for some c ∈ [a , b].
First we consider c ∈ (a , b).
Let ε > 0 be given.
f 0 (x)
Since lim 0
= L,
x→c g (x)
∃ δ > 0 such that
f 0 (x)
− L < ε for all x ∈ (c−δ , c)∪(c , c+δ). (2.14)
g 0 (x)
Case(i): x ∈ (c − δ , c).
By the Cauchy Mean Value Theorem,
f (c) − f (x)
f 0 (cx )
∃ cx ∈ (x , c) such that
= 0
.
g(c) − g(x)
g (cx )
f (x)
f 0 (cx )
i.e.
= 0
.
g(x)
g (cx )
Since cx ∈ (c − δ , c), from (4.14), we have,
f (x)
f 0 (cx )
−L <
− L < ε for all x ∈ (c − δ , c).
g(x)
g 0 (cx )
Case(ii) x ∈ (c , c + δ).
By the Cauchy Mean Value Theorem,
f (c) − f (x)
f 0 (dx )
∃ dx ∈ (c , x) such that
= 0
.
g(c) − g(x)
g (dx )
f (x)
f 0 (dx )
= 0
.
i.e.
g(x)
g (dx )
Since dx ∈ (c , c + δ) from (4.15), we have,
61
(2.15)
f (x)
f 0 (dx )
−L <
−L < ε .
g(x)
g 0 (dx )
Since x ∈ (c , c + δ) is arbitrary,
∃ δ > 0 such that
f (x)
− L < ε for all x ∈ (c , c + δ).
g(x)
(2.16)
From (4.15) & (4.16), we have,
f (x)
− L < ε for all x ∈ (c − δ , c) ∪ (c , c + δ).
∃ δ > 0 such that
g(x)
Since ε > 0 is arbitrary,
f (x)
∀ ε > 0, ∃ δ > 0 such that
− L < ε for all x ∈ (c−δ , c)∪(c , c+δ).
g(x)
Hence by the definition,
f (x)
= L.
lim
x→c g(x)
For the cases c = a and c = b, the proof is similar to the case(ii) and
case(i) respectively.
(2). Exercise.
Eg:
1.
sin x
lim+ √
x→0
x
√
sin x
cos x
lim+ √ = lim+ 1 = lim+ 2 x cos x = 0.
√
x→0
x→0
x→0
x
2 x
(1 − cos x)
sin x
cos x
1
= lim
= lim
= .
2
x→0
x→0 2x
x→0
x
2
2
2. lim
ln x
1
= lim = 1.
x→1 x
x→1 (x − 1)
3. lim
L’Hospital’s Rule: The case
∞
.
∞
Theorem 2.14. Suppose that f and g are differentiable on (a , b) and
continuous on [a , b] and that lim f (x) = ∞ and lim g(x) = ∞ for some
x→c
x→c
c ∈ [a , b] and that g(x) 6= 0 and g 0 (x) 6= 0 for all x ∈ [a , b] \ {c}. Then
f 0 (x)
f (x)
= L (∈ R), then lim
= L;
0
x→c g (x)
x→c g(x)
1. if lim
f 0 (x)
f (x)
= +∞ (or − ∞), then lim
= +∞ (or − ∞).
0
x→c g (x)
x→c g(x)
2. if lim
Eg:
62
1
ln x
= lim x = 0.
x→∞ 1
x→∞ x
1. lim
ln sin x
2. lim+
= lim+
x→0
x→0
ln x
cos x
sin x
1
x
x · cos x = 1.
= lim+
x→0
sin x
Examples of other indeterminate forms.
Eg:
1. lim+
x→0
1
1
−
x sin x
(∞ − ∞ form).
1
1
lim
−
x→0+
x sin x
(sin x − x)
= lim+
.
x→0
x sin x
=
lim+
(cos x − 1)
.
sin x + x cos x
lim+
− sin x
.
2 cos x − x sin x
x→0
=
x→0
=
0
= 0.
2
(0 · (−∞) form).
2. lim+ x ln x
x→0
lim x ln x
x→0+
=
lim+
x→0
=
lim+
x→0
=
ln x
1
x
.
1/x
.
−1/x2
lim (−x) = 0.
x→0+
3. lim+ xx
(00 form).
x→0
It is from the calculus that,
xx = ex ln x .
63
Since exponential function is continuous,
lim+ xx = lim+ ex ln x .
x→0
x→0
lim x ln x
= ex→0+
.
= e0 = 1.
4. lim
x→∞
1
1+
x
x
(1∞ form).
We
notethat,
x
1
1
1+
= ex ln(1+ x ) .
x
Further, we have, 1
lim x ln 1 +
=
x→∞
x
=
=
ln 1 +
lim
x→∞
1/x
lim
1+
x→∞
1
x→∞ 1 +
lim
Therefore
lim
x→∞
1
1+
x
x
1
x
1
x
.
1 −1
(−x−2 )
x
.
−x−2
= 1.
1
lim ex ln(1+ x ) .
x→∞
1
lim x ln 1 +
x .
= ex→∞
=
= e1 = e.
64