4.1 – 4.

3 The Mean Value Theorem

Theorem (The Extreme Value Theorem)
If f is continuous on [a, b], then f attains an absolute maximum value f (c) and
an absolute minimum value f (d) at some numbers c and d in [a, b].

Theorem (Fermat’s Theorem)

If f has a local maximum or minimum at c, and if f → (c) exists, then

f → (c) = 0.

Theorem (Rolle’s Theorem)

If f is continuous on [a, b] and di!erentiable on (a, b) and f (a) = f (b), then
there is a number c in (a, b) such that f → (c) = 0.

1
Example
Prove that the equation x3 + x → 1 = 0 has exactly one real solution.

Solution. Let f (x) = x3 + x → 1. Then f (0) = →1 < 0, f (1) = 1 > 0 and f is

continuous on [0, 1]. By IVT, there is k ↑ (0, 1) such that f (k) = 0.
Thus the given equation has a solution.

Suppose it has two or more real solutions.

Choose two distinct solutions a and b with a < b. Then f (a) = 0 = f (b) and f
is continuous on [a, b] and di!erentiable on (a, b). By Rolle’s theorem, there is
c ↑ (a, b) such that f → (c) = 0. But f → (x) = 3x2 + 1 ↓ 1 for all x ↑ R and so
f → (x) ↔= 0 for all x ↑ R. This gives a contradiction.
Therefore the equation can not have two or more real solutions.

2
Theorem (The Mean Value Theorem)
If f is continuous on [a, b] and di!erentiable on (a, b), then there is a number c
in (a, b) such that
f (b) → f (a)
f → (c) =
b→a
or, equivalently,
f (b) → f (a) = f → (c)(b → a)

3
Theorem
If f → (x) = 0 for all x ↑ I, then f is constant on I.

Proof.
Suppose x1 , x2 ↑ I with x1 < x2 .
Then f is di!erentiable on (x1 , x2 ) and continuous on [x1 , x2 ].
By MVT, there is c ↑ (x1 , x2 ) such that
f (x2 ) → f (x1 )
= f → (c)
x2 → x1
Since f → (c) = 0,

f (x2 ) → f (x1 )
= 0 ↗ f (x2 ) → f (x1 ) = 0 ↗ f (x1 ) = f (x2 )
x2 → x1
Therefore f has the same value at any two numbers x1 and x2 in I.
This means that f is constant on I.

4
Corollary
If f → (x) = g → (x) for all x ↑ I, then f → g is constant on I; that is, there is a
constant C such that

f (x) = g(x) + C for all x ↑ I.

Proof.
Let F (x) = f (x) → g(x). Then

F → (x) = f → (x) → g → (x) = 0 for all x ↑ I.

Thus F = f → g is constant on I, that is, there is a constant C such that

f (x) = g(x) + C for all x ↑ I.

5
4.4 Indeterminate forms and L’Hospital’s Rule
If we have a limit of the form
f (x)
lim
x↑a g(x)
where both f (x) ↘ 0 and g(x) ↘ 0 as x ↘ a, then this limit may or may not
exist and is called an indeterminate form of type 00 .

If we have a limit of the form

f (x)
lim
x↑a g(x)
where both f (x) ↘ ≃(or → ≃) and g(x) ↘ ≃(or → ≃) as x ↘ a, then this
limit may or may not exist and is called an indeterminate form of type ↓
↓
.

6
Theorem (L’Hospital’s Rule)
Suppose f and g are di!erentiable and g → (x) ↔= 0 on an open interval I that
contains a (except possibly at a).
Suppose that
lim f (x) = 0 and lim g(x) = 0
x↑a x↑a

or that
lim f (x) = ±≃ and lim g(x) = ±≃
x↑a x↑a
0 ↓
(In other word, we have an indeterminate form of type 0
or ↓
.) Then

f (x) f → (x)
lim = lim →
x↑a g(x) x↑a g (x)

if the limit on the right side exists (or is ≃ or →≃).

Remark
L’Hospital’s Rule is also valid for one-sided limits and for limits at infinity or
negative infinity; That is, ‘x ↘ a’ can be replaced by any of the symbols
x ↘ a+ , x ↘ a↔ , x ↘ ≃, or x ↘ →≃.

7
Example
ln x
Find lim .
x↑1 x→1
Solution.
Let f (x) = ln x and g(x) = x → 1. Then f and g are di!erentiable and
g → (x) ↔= 0 on an open interval I that contains 1 and

lim f (x) = ln 1 = 0 and lim g(x) = 0.

x↑1 x↑1

Thus we can apply l’Hospital’s Rule:

d
ln x H dx
(ln x) 1/x 1
lim = lim d
= lim = lim = 1
x↑1 x→1 x↑1
dx
(x → 1) x↑1 1 x↑1 x

8
Example
ex
Calculate lim .
x↑↓ x2
Solution.

ex H ex H ex
lim 2
= lim = lim =≃
x↑↓ x x↑↓ 2x x↑↓ 2

⇐ ⇐
type ↓
↓
type ↓
↓

Example
ln x
Calculate lim ⇒ .
x↑↓ 3 x

Solution.

ln x H 1/x 1/x 2
lim ⇒ = lim 1 ↔1/2 = lim ⇒ = lim ⇒ = 0
x↑↓ x x↑↓ x
2
x↑↓ 1/2 x x↑↓ x
⇐
↓
type ↓

9
Example
tan x → x
Find lim .
x↑0 x3
Solution.
tan x → x H sec3 x → 1 H 2 sec2 x tan x
lim 3
= lim 2
= lim
x↑0 x x↑0 3x x↑0 6x
⇐ ⇐
0 0
type 0
type 0

tan x → x 2 sec2 x tan x 1 tan x

lim = lim = lim sec2 x ·
x↑0 x3 x↑0 6x x↑0 3 x
1 2 sin x 1 1 2 sin x 1
= lim sec x · · = lim sec x · lim · lim
x↑0 3 x cos x 3 x↑0 x↑0 x x↑0 cos x
1 1
= ·1·1·1=
3 3

10
Example
sin x
Find lim .
x↑ω↔ 1 → cos x
Solution.
sin x sin ω 0
lim = = =0
x↑ω↔ 1 → cos x 1 → cos ω 1 → (→1)

11
• Indeterminate products

If lim f (x) = 0 and lim g(x) = ≃ (or →≃), then the limit
x↑a x↑a

lim f (x)g(x)
x↑a

is called an indeterminate form of type 0 · ≃.

Example
Evaluate lim x ln x.
x↑0+
Solution.
The given limit is indeterminate form of type 0 · ≃.
ln x H 1/x
lim x ln x = lim = lim 2
= lim (→x) = 0
x↑0+ x↑0+ 1/x x↑0+ →1/x x↑0+

⇐ ⇐
↓
type 0 · ≃ type ↓

12
• Indeterminate Di!erences

If lim f (x) = ≃ and lim g(x) = ≃, then the limit

x↑a x↑a

lim [f (x) → g(x)]

x↑a

is called an indeterminate form of type ≃ → ≃.

Example
Compute lim
ω
(sec x → tan x).
x↑ 2 ↔

Solution. The given limit is indeterminate form of type ≃ → ≃.

1 → sin x H → cos x 0
lim (sec x → tan x) = lim = lim = lim =0
x↑ ω
2
↔ ω
x↑ 2 ↔ cos x x↑ ω
2
↔ → sin x x↑ ω ↔ →1
2

⇐ ⇐
0
type ≃ → ≃ type 0

13
• Indeterminate Powers

Several indeterminate forms arise from the limit

lim [f (x)]g(x)
x↑a

(1) lim f (x) = 0 and lim g(x) = 0 type 00

x↑a x↑a

(2) lim f (x) = ≃ and lim g(x) = 0 type ≃0

x↑a x↑a
(3) lim f (x) = 1 and lim g(x) = ±≃ type 1↓
x↑a x↑a

14
Example
Calculate lim (1 + sin 4x)cot x .
x↑0+
Solution.
The given limit is indeterminate form of type 1↓ . Let

y = (1 + sin 4x)cot x

Then
ln(1 + sin 4x)
ln y = ln[(1 + sin 4x)cot x ] = cot x ln(1 + sin 4x) =
tan x
By l’Hospital’s Rule,
4 cos 4x
ln(1 + sin 4x) H
lim ln y = lim = lim 1+sin2 4x = 4
x↑0+ x↑0+ tan x x↑0+ sec x

⇐
type 00
Since limx↑0+ ln y = 4 and y = ex is continuous at 4, we have

lim (1 + sin 4x)cot x = lim eln y = elimx→0+ ln y = e4

x↑0+ x↑0+

15
Example
Find lim xx .
x↑0+
Solution.
The given limit is indeterminate form of type 00 . Note that

lim x ln x = 0
x↑0+

Since lim x ln x = 4 and y = ex is continuous at 0, we have

x↑0+

lim xx = lim ex ln x = elimx→0+ x ln x = e0 = 1

x↑0+ x↑0+

16
Theorem (Cauchy’s Mean Value Theorem)
Suppose that f and g are continuous on [a, b] and di!erentiable on (a, b), and
g → (x) ↔= 0 for all x in (a, b). Then there is a number c in (a, b) such that

f → (c) f (b) → f (a)

=
g → (c) g(b) → g(a)

Proof. First we show that g(a) ↔= g(b). If g(a) = g(b), then, by MVT, there is
a number x0 ↑ (a, b) such that

g(b) → g(a)
0= = g → (x0 .)
b→a
This cannot happen because g → (x) ↔= 0 for all x ↑ (a, b).
Let
f (b) → f (a) ! "
F (x) = f (x) → f (a) → g(x) → g(a) .
g(b) → g(a)
Then F is continuous on [a, b] and di!erential on (a, b). Also F (b) = F (a). By
MVT again, there is a number c ↑ (a, b) such that

F (b) → F (a)
0= = F → (c).
b→a

17
But
f (b) → f (a) →
F → (x) = f → (x) → g (x).
g(b) → g(a)
Hence
f (b) → f (a) →
0 = F → (c) = f → (c) → g (c).
g(b) → g(a)
Since g → (c) ↔= 0,
f → (c) f (b) → f (a)
= .
g → (c) g(b) → g(a)

Remark
If we take the special case in which g(x) = x, then g → (c) = 1 and Cauchy’s
MVT is just the ordinary MVT.

18
Proof of l’Hospital’s Rule.
We are assuming that limx↑a f (x) = 0 and limx↑a g(x) = 0.
Let
f → (x)
L = lim → .
x↑a g (x)

We must show that

f (x)
lim = L.
x↑a g(x)
Define
# #
f (x) if x ↔= a g(x) if x ↔= a
F (x) = and G(x) =
0 if x = a 0 if x = a.

Then F is continuous on I since f is continuous on {x ↑ I | x ↔= a} and

lim F (x) = lim f (x) = 0 = F (a) .
x↑a x↑a

Likewise, G is continuous on I. Let x ↑ I and x > a. Then F and G are

continuous on [a, x] and di!erentiable on (a, x) and G→ (t) ↔= 0 for all t ↑ (a, x)
(since G→ (t) = g → (t) ↔= 0 for all t ↑ (a, x)). Therefore, by Cauchy’s MVT, there
is a number y ↑ (a, x) such that
F → (y) F (x) → F (a) F (x)
= =
G→ (y) G (x) → G (a) G (x)
19
Here, we have used the fact that, by definition, F (a) = 0 and G(a) = 0. Now,
if we let x ↘ a+ , then y ↘ a+ (since a < y < x), so

f (x) F (x) F → (y) f → (y)

lim = lim = lim →
= lim → =L
x↑a+ g(x) x↑a+ G(x) y↑a+ G (y) y↑a+ g (y)

A similar argument shows that the left-hand limit is also L. Therefore

f (x)
lim = L.
x↑a g (x)

This proves l’Hospital’s Rule for the case where a is finite.

If a is infinite, we let t = x1 Then t ↘ 0+ as x ↘ ≃, we have

f (x) f (1/t)
lim = lim
x↑↓ g(x) t↑0+ g(1/t)

f → (1/t)(→1/t2 )
= lim
t↑0+g → (1/t)(→1/t2 )
f → (1/t) f → (x)
= lim → = lim →
t↑0+ g (1/t) x↑↓ g (x)

20
G. Concavity and points of inflection

f → (x) = xex + ex = (x + 1)ex

f →→ (x) = (x + 1)ex + ex = (x + 2)ex

x ··· →2 ··· →1 ···

f ↑ (x) → → → 0 +
f ↑↑ (x) → 0 + + +
f (x) ↑ → e22 ↑ → 1e ↓
CD IP CU LMIN CD

H. Sketch the curve

22
Example
2x2
Sketch the curve f (x) = .
x2→1

Solution.
A. The domain is
{x | x2 → 1 ↔= 0} = {x | x ↔= ±1} = (→≃, →1) ⇑ (→1, 1) ⇑ (1, ≃)
B. The x- and y-intercepts are both 0.
C. Since f (→x) = f (x), f is even. The curve is symmetric about the y-axis.
D.
2x2 2
lim = lim = =2
x↑±↓ x2 → 1 x↑±↓ 1 → 1/x2
Therefore the line y = 2 is a horizontal asymptote.
Since the denominator is 0 when x = ±1, we compute the following limits:
2x2 2x2
lim =≃ lim = →≃
x↑1+ x2 → 1 2
x↑1↓ x → 1

2x2 2x2
lim 2
= →≃ lim 2
=≃
x↑↔1 x → 1
+ x↑↔1 x → 1
↓

Therefore the lines x = 1 and x = →1 are vertical asymptotes.

E. Intervals of increase or decrease
F. Local maximum and minimum values
23
G. Concavity and points of inflection
4x(x2 → 1) → 2x2 · 2x →4x
f → (x) = =
(x2 → 1)2 (x2 → 1)2
2
→4(x2 → 1) + 4x · 2(x2 → 1)2x 12x2 + 4
f →→ (x) = 4 =
(x2 → 1) (x2 → 1)3

x ··· →1 ··· 0 ··· 1 ···

f ↑ (x) + ↔ + 0 → ↔ →
f ↑↑ (x) + ↔ → → → ↔ +
f (x) ↓ ↔ ↓ 0 ↑ ↔ ↑
CU CD LMAX CD CU

H. Sketch the curve

24
Definition
If

lim [f (x) → (mx + b)] = 0

x↑↓

or lim [f (x) → (mx + b)] = 0

x↑↔↓

then the line y = mx + b is called a slant asymptote.

25
Example
x3
Sketch the graph of f (x) = .
x2 +1

Solution.
A. The domain is R = (→≃, ≃).
B. The x- and y-intercepts are both 0.
C. As f (→x) = →f (x), f is odd and its graph is symmetric about the origin.
D. Since x2 + 1 is never 0, there is no vertical asymptote.
Since f (x) ↘ ≃ as x ↘ ≃ and f (x) ↘ →≃ as x ↘ →≃, there is no
horizontal asymptote. But

x3 x
f (x) = =x→ 2
x2 +1 x +1
1
x
f (x) → x = → = → x 1 ↘ 0 as x ↘ ±≃
x2 + 1 1 + x2
So the line y = x is a slant asymptote.
E. Intervals of increase or decrease
F. Local maximum and minimum values

26
G. Concavity and points of inflection
3x2 (x2 + 1) → x3 · 2x x2 (x2 + 3)
f → (x) = 2 =
(x2 + 1) (x2 + 1)2
2
(4x3 + 6x)(x2 + 1) → (x4 + 3x2 ) · 2(x2 + 1)2x 2x(3 → x2 )
f →→ (x) = 4 =
(x2 + 1) (x2 + 1)3
↗ ↗
x ··· → 3 ··· 0 ··· 3 ···
f ↑ (x) + + + 0 + + +
f ↑↑ (x) + 0 → 0 + 0 →
↔ ↔
f (x) ↓ →343 ↓ 0 ↓ 3
4
3
↓
CU IP CD IP CU IP CD

H. Sketch the curve

27