Section 4.5L’Hôpital’s Rule and IndeterminateForms

(1) Refresher: Determinate and Indeterminate Forms(2) L’Hôpital’s Rule(3) Comparing Growth of Functions

The Form of a Limit

The form of a limit limx→c

� is the expression resulting from substitutingx = c into �.

The form of a limit is not the same as its value!It is a tool for inspecting the limit.

limx→0

xarctan(x) : form 00 limx→∞(1+x)

1x : form ∞0

limx→0

cos(x)1x : form 1∞ lim

x→0+ ln(x)sin(x) : form 0 ·∞

Indeterminate Forms are called indeterminate because theyrepresent limits which may or may not exist and may be equal to anyvalue. The form itself does not indicate the value of the limit. There are7 indeterminate forms.

00

±∞∞ ±0 ·∞

1∞ 00 ∞0

∞−∞

Indeterminate Forms

Limits of form00:

limx→0

sin(x)

x= 1 lim

x→0

sin2(x)

x= 0 lim

x→0

sin(x)

x2 DNE

Limits of form∞∞ :

limx→∞

x +1x

= 1 limx→∞

ln(x)

x= 0 lim

x→∞ex

x=∞

These two indeterminate forms are like “tugs-of-war” between thenumerator and denominator. Which of the two grows faster?

L’Hôpital’s Rule

L’Hôpital’s RuleIf f and g are differentiable near x = a and either

(i) limx→a

f (x)= 0 and limx→a

g(x)= 0, or

(ii) limx→a

f (x)=±∞ and limx→a

g(x)=±∞,

then

limx→a

f (x)

g(x)= lim

x→a

f ′(x)g ′(x)

.

The rule applies equally for one-sided limits.

Example 1:

These limits are of form 0/0. The steps marked LHR= use L’Hôpital’s Rule.

(a) limx→0

sin(x)

xLHR= lim

x→0

ddx (sin(x))

ddx (x)

= limx→0

cos(x)

1= cos(0) = 1

(b) limx→0

ln(x2+1)x

LHR= limx→0

2xx2+11

= 0

(c) limx→2

x4+2x −20x3−8

LHR= limx→2

4x3+23x2 = 17

6

Example 2:

These limits are of form ∞∞ . The steps marked LHR= use L’Hôpital’s Rule.

(a) limx→∞

3x −76x +5

LHR= limx→∞

ddx (3x −7)ddx (6x +5)

= limx→∞

36

= 12

(b) limx→∞

ln(x)

xLHR= lim

x→∞

ddx (ln(x))

ddx (x)

= limx→∞

1/x1

= 0

(c) limx→0+

ln(x)

ln(sin(x))LHR= lim

x→0+

1x

cos(x)sin(x)

= limx→0+

(sin(x)

x

)(1

cos(x)

)= 1

Example 3:Sometimes it is necessary to perform L’Hôpital’s Rule multiple times.

(a) limx→0

ex −x −1cos(x)−1

(0/0) = limx→0

ex −1−sin(x)

(0/0)

= limx→0

ex

−cos(x)(determinate) = −1.

(b) limx→∞

−3x3−20x2

2x3+x +7(∞/∞) = lim

x→∞−9x2−40x6x2+1

(∞/∞)

= limx→∞

−18x −4012x

(∞/∞)

= limx→∞

−1812

= −32

.

L’Hôpital’s Rule: Warnings

Warning #1:L’Hôpital’s Rule only applies to the indeterminate forms 0/0 and ∞/∞.Before applying L’Hôpital’s Rule to a limit, verify that it is of one ofthose two forms.

Warning #2:Don’t confuse L’Hôpital’s Rule with the Quotient Rule!They have completely different uses.

L’Hôpital’s Rule is for evaluating limits.The Quotient Rule is for evaluating derivatives.

The Form ∞−∞To evaluate limits with the form ∞−∞, use algebra to rewrite theexpression as a quotient, often in 0/0 or ∞/∞ form. Often this meansfinding a common denominator.

Example 4: To evaluate limx→0+

(1x−csc(x)

), which has the form ∞−∞,

rewrite it:

1x−csc(x) = 1

x− 1sin(x)

= sin(x)−x

x sin(x)

This is an 0/0 form, so we can apply L’Hôpital’s Rule:

limx→0+

sin(x)−x

x sin(x)LHR= lim

x→0+cos(x)−1

x cos(x)+ sin(x)(0/0)

LHR= limx→0+

−sin(x)

−x sin(x)+2cos(x)= 0

The Form 0 ·∞

Limits with the form 0 ·∞ can easily be converted to 0/0 or ∞/∞ form.

If limx→a f (x)= 0 and limx→a g(x)=∞, then

limx→a

f (x)g(x)︸ ︷︷ ︸0·∞ form

= limx→a

f (x)(1

g(x)

)︸ ︷︷ ︸0/0 form

= limx→a

g(x)(1

f (x)

)︸ ︷︷ ︸∞/∞ form

Examples (5):

(a) limx→0+ x ln(x) = lim

x→0+ln(x)

1x

LHR= limx→0+

1x

− 1x2

= 0

(b) limx→ π

2−

(x − π

2

)tan(x)= lim

x→ π2−x − π

2cot(x)

LHR= limx→ π

2−

1−csc2(x)

= −1

Indeterminate Forms Involving Exponents

For limits of the forms 1∞, 00, or ∞0, the key step is to rewrite

limx→c

■ = limx→c

eln(■) = limx→c

exp(ln(■)) = exp(limx→c

ln(■))

.

The second equality is valid by the Limit Laws (see §2.4) because ex

is continuous everywhere.The “exp” notation avoids humongous expressions in superscript.

This technique changes the limit into one of form 0 ·∞:

1∞ → e ln|1∞| → e∞·ln(1) ↘

00 → e ln∣∣00∣∣ → e0·ln(0) → e0·∞

∞0 → e ln∣∣∞0∣∣ → e0·ln(∞) ↗

Indeterminate Forms Involving Exponents

Example 6: limx→0+ x

x has the indeterminate form 00.

limx→0+ x

x = exp

(limx→0+ ln(x

x )

)= exp

(limx→0+ x ln(x)

)(0 ·∞)

= exp

(limx→0+

ln(x)1x

)(∞/∞)

LHR= exp

(limx→0+

1/x−1/x2

)= exp

(limx→0+(−x)

)

= exp(0)= 1.

Indeterminate Forms Involving Exponents

Example 7: limx→0+(− ln(x))x has the indeterminate form ∞0.

limx→0+(− ln(x))x = exp

[limx→0+ ln((− ln(x))x )

]= exp

[limx→0+ x ln(− ln(x))

]= exp

[limx→0+

ln(− ln(x))

1/x

](∞/∞)

LHR= exp

limx→0+

1− ln(x) · −1

x

−1x2

= exp

[limx→0+

−xln(x)

]

= exp(0)= 1.

Indeterminate Forms Involving Exponents

Example 8: Let a be any real number, so that limx→0

(1+ax)1/x has the

indeterminate form 1∞.

limx→0

(1+ax)1/x = exp

[limx→0+ ln

((1+ax)1/x

)]= exp

[limx→0+

ln(1+ax)

x

](0/0)

LHR= exp

[limx→0+

a1+ax1

]= ea.

In fact, this calculation can be used to define the number e:

e = limx→0

(1+x)1/x .

Comparing Growth Of Functions

We say that f grows faster than g if

(i) limx→∞

f (x)

g(x)=∞, or equivalently

(ii) limx→∞

g(x)

f (x)= 0.

(Notation: g ¿ f .)

x¿ x2

x

y

2 4 6 8

10

20

30

40

y = x2y = x

Some important cases (that can all be checked using L’Hôpital’s Rule):

(a) xn ¿ ex for all n.(b) In fact, xn ¿ ax for all n and all a> 1.(c) loga(x)¿ xn for all n and all a> 0.

Indeterminate Forms Involving TrigonometricFunctions

Example 9: limx→0tan(x)sin(5x)

x sin(7x)has the indeterminate form

00.

limx→0

tan(x)sin(5x)x sin(7x)

= limx→0

[(1

cos(x)

)(sin(x)

x

)(sin(5x)sin(7x)

)]Limit of each fraction exists.

=���

���*

1

limx→0

(1

cos(x)

)limx→0

(sin(x)

x

)limx→0

(sin(5x)sin(7x)

)LHR= lim

x→0

(cos(x)

1

)limx→0

(5cos(5x)7cos(7x)

)= 57