Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 021 slides)

Section 3.7Indeterminate Forms and L’Hôpital’s

V63.0121.021, Calculus I

New York University

November 4, 2010

Announcements

I Quiz 3 in recitation this week on Sections 2.6, 2.8, 3.1, and 3.2

. . . . . .

Announcements

I Quiz 3 in recitation thisweek on Sections 2.6, 2.8,3.1, and 3.2

V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 2 / 35

. . . . . .

Objectives

I Know when a limit is ofindeterminate form:

I indeterminate quotients:0/0, ∞/∞

I indeterminate products:0×∞

I indeterminatedifferences: ∞−∞

I indeterminate powers:00, ∞0, and 1∞

I Resolve limits inindeterminate form

. . . . . .

Recall

Recall the limit laws from Chapter 2.

I Limit of a sum is the sum of the limits

I Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI Limit of a quotient is the quotient of the limits ... whoops! This is

true as long as you don’t try to divide by zero.

. . . . . .

Recall

I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limits

I Limit of a product is the product of the limitsI Limit of a quotient is the quotient of the limits ... whoops! This is

. . . . . .

Recall

I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limits

I Limit of a quotient is the quotient of the limits ... whoops! This istrue as long as you don’t try to divide by zero.

. . . . . .

Recall

I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI Limit of a quotient is the quotient of the limits ... whoops! This is

. . . . . .

More about dividing limits

I We know dividing by zero is bad.I Most of the time, if an expression’s numerator approaches a finite

nonzero number and denominator approaches zero, the quotienthas an infinite. For example:

limx→0+

1x= +∞ lim

x→0−cos xx3

= −∞

. . . . . .

Why 1/0 ̸= ∞

Consider the function f(x) =1

1x sin x

Then limx→∞

f(x) is of the form 1/0, but the limit does not exist and is notinfinite.

Even less predictable: when numerator and denominator both go tozero.

. . . . . .

Why 1/0 ̸= ∞

Consider the function f(x) =1

1x sin x

Then limx→∞

f(x) is of the form 1/0, but the limit does not exist and is notinfinite.Even less predictable: when numerator and denominator both go tozero.

. . . . . .

Experiments with funny limits

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

All of these are of the form00, and since we can get different answers

in different cases, we say this form is indeterminate.

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

I limx→0

sin2 xx

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

I limx→0

sin 3xsin x

. . . . . .

Language NoteIt depends on what the meaning of the word “is" is

I Be careful with thelanguage here. We are notsaying that the limit in each

case “is”00, and therefore

nonexistent because thisexpression is undefined.

I The limit is of the form00,

which means we cannotevaluate it with our limitlaws.

. . . . . .

Indeterminate forms are like Tug Of War

Which side wins depends on which side is stronger.

. . . . . .

Outline

L’Hôpital’s Rule

Relative Rates of Growth

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

. . . . . .

The Linear Case

QuestionIf f and g are lines and f(a) = g(a) = 0, what is

limx→a

f(x)g(x)

SolutionThe functions f and g can be written in the form

f(x) = m1(x− a)g(x) = m2(x− a)

Sof(x)g(x)

. . . . . .

The Linear Case

QuestionIf f and g are lines and f(a) = g(a) = 0, what is

limx→a

f(x)g(x)

SolutionThe functions f and g can be written in the form

f(x) = m1(x− a)g(x) = m2(x− a)

Sof(x)g(x)

. . . . . .

The Linear Case, Illustrated

.y = f(x)

.y = g(x)

.f(x) .g(x)

f(x)g(x)

=f(x)− f(a)g(x)− g(a)

=(f(x)− f(a))/(x− a)(g(x)− g(a))/(x− a)

. . . . . .

What then?

I But what if the functions aren’t linear?

I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function? The derivative!

. . . . . .

What then?

I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?

I What would be the slope of that linear function? The derivative!

. . . . . .

What then?

I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function?

The derivative!

. . . . . .

What then?

I But what if the functions aren’t linear?I Can we approximate a function near a point with a linear function?I What would be the slope of that linear function? The derivative!

. . . . . .

Theorem of the Day

Theorem (L’Hopital’s Rule)

Suppose f and g are differentiable functions and g′(x) ̸= 0 near a(except possibly at a). Suppose that

limx→a

f(x) = 0 and limx→a

g(x) = 0

limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞

Thenlimx→a

f(x)g(x)

= limx→a

f′(x)g′(x)

if the limit on the right-hand side is finite, ∞, or −∞.

. . . . . .

Meet the Mathematician

I wanted to be a militaryman, but poor eyesightforced him into math

I did some math on his own(solved the “brachistocroneproblem”)

I paid a stipend to JohannBernoulli, who proved thistheorem and named it afterhim! Guillaume François Antoine,

Marquis de L’Hôpital(French, 1661–1704)

. . . . . .

Revisiting the previous examples

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

cos x2 − 2x2 sin(x2)

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x.

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x.

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x.

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x.

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x.

.numerator → 0

sin x2.

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x.

.numerator → 0

sin x2.

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x.

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x.

.numerator → 0

(cos x2

)(�2x.

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x.

.numerator → 0

(cos x2

)(�2x.

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x.

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x.

.numerator → 1

cos x2 − 2x2 sin(x2).

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Example

limx→0

sin2 xx

H= lim

2 sin x

.sin x → 0

cos x1

Example

limx→0

sin2 x

.numerator → 0

sin x2

.denominator → 0

H= lim

�2 sin x cos x

.numerator → 0

(cos x2

)(�2x

.denominator → 0

H= lim

cos2 x− sin2 x

.numerator → 1

.denominator → 1

Example

limx→0

sin 3xsin x

H= lim

3 cos 3xcos x

. . . . . .

Another Example

Example

Findlimx→0

xcos x

SolutionThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.

. . . . . .

Beware of Red Herrings

Example

Findlimx→0

xcos x

SolutionThe limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.

. . . . . .

Outline

. . . . . .

Limits of Rational Functions revisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

SolutionUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

. . . . . .

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

. . . . . .

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

. . . . . .

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

. . . . . .

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

. . . . . .

Limits of Rational Functions revisited II

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

. . . . . .

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

. . . . . .

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

. . . . . .

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

. . . . . .

Limits of Rational Functions revisited III

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

. . . . . .

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

. . . . . .

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

. . . . . .

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

. . . . . .

Limits of Rational Functions

FactLet f(x) and g(x) be polynomials of degree p and q.

I If p > q, then limx→∞

f(x)g(x)

I If p < q, then limx→∞

f(x)g(x)

I If p = q, then limx→∞

f(x)g(x)

is the ratio of the leading coefficients of f

and g.

. . . . . .

Exponential versus geometric growth

Example

Find limx→∞

x2, if it exists.

SolutionWe have

limx→∞

x2H= lim

x→∞

2xH= lim

x→∞

2= ∞.

Example

What about limx→∞

AnswerStill ∞. (Why?)

. . . . . .

Example

Find limx→∞

x2, if it exists.

SolutionWe have

limx→∞

x2H= lim

x→∞

2xH= lim

x→∞

2= ∞.

Example

. . . . . .

Example

Find limx→∞

x2, if it exists.

SolutionWe have

limx→∞

x2H= lim

x→∞

2xH= lim

x→∞

2= ∞.

Example

. . . . . .

Example

Find limx→∞

x2, if it exists.

SolutionWe have

limx→∞

x2H= lim

x→∞

2xH= lim

x→∞

2= ∞.

Example

. . . . . .

Exponential versus fractional powers

Example

Find limx→∞

ex√x, if it exists.

Solution (without L’Hôpital)

We have for all x > 1, x1/2 < x1, so

The right hand side tends to ∞, so the left-hand side must, too.

Solution (with L’Hôpital)

limx→∞

ex√x= lim

x→∞

12x−1/2

= limx→∞

2√xex = ∞

. . . . . .

Example

Find limx→∞

limx→∞

ex√x= lim

x→∞

12x−1/2

= limx→∞

2√xex = ∞

. . . . . .

Example

Find limx→∞

limx→∞

ex√x= lim

x→∞

12x−1/2

= limx→∞

2√xex = ∞

. . . . . .

Exponential versus any power

TheoremLet r be any positive number. Then

limx→∞

xr= ∞.

Proof.If r is a positive integer, then apply L’Hôpital’s rule r times to thefraction. You get

limx→∞

xrH= . . .

H= lim

x→∞

r!= ∞.

If r is not an integer, let m be the smallest integer greater than r. Then

if x > 1, xr < xm, soex

xm. The right-hand side tends to ∞ by the

previous step.

. . . . . .

Exponential versus any power

limx→∞

xr= ∞.

Proof.If r is a positive integer, then apply L’Hôpital’s rule r times to thefraction. You get

limx→∞

xrH= . . .

H= lim

x→∞

r!= ∞.

If r is not an integer, let m be the smallest integer greater than r. Then

if x > 1, xr < xm, soex

xm. The right-hand side tends to ∞ by the

previous step.V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 25 / 35

. . . . . .

Any exponential versus any power

TheoremLet a > 1 and r > 0. Then

limx→∞

xr= ∞.

Proof.If r is a positive integer, we have

limx→∞

xrH= . . .

H= lim

x→∞

(lna)rax

r!= ∞.

If r isn’t an integer, we can compare it as before.

So even limx→∞

(1.00000001)x

x100000000= ∞!

. . . . . .

limx→∞

xr= ∞.

limx→∞

xrH= . . .

H= lim

x→∞

(lna)rax

r!= ∞.

So even limx→∞

(1.00000001)x

x100000000= ∞!

. . . . . .

limx→∞

xr= ∞.

limx→∞

xrH= . . .

H= lim

x→∞

(lna)rax

r!= ∞.

So even limx→∞

(1.00000001)x

x100000000= ∞!

. . . . . .

Logarithmic versus power growth

limx→∞

ln xxr

Proof.One application of L’Hôpital’s Rule here suffices:

limx→∞

ln xxr

H= lim

x→∞

1/xrxr−1 = lim

x→∞

. . . . . .

Logarithmic versus power growth

limx→∞

ln xxr

Proof.One application of L’Hôpital’s Rule here suffices:

limx→∞

ln xxr

H= lim

x→∞

1/xrxr−1 = lim

x→∞

. . . . . .

Outline

. . . . . .

Indeterminate products

Example

Findlim

x→0+

√x ln x

This limit is of the form 0 · (−∞).

SolutionJury-rig the expression to make an indeterminate quotient. Then applyL’Hôpital’s Rule:

limx→0+

√x ln x

= limx→0+

ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x = 0

. . . . . .

Example

Findlim

x→0+

√x ln x

limx→0+

√x ln x

= limx→0+

ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x = 0

. . . . . .

Example

Findlim

x→0+

√x ln x

limx→0+

√x ln x = lim

x→0+ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x = 0

. . . . . .

Example

Findlim

x→0+

√x ln x

limx→0+

√x ln x = lim

x→0+ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x = 0

. . . . . .

Example

Findlim

x→0+

√x ln x

limx→0+

√x ln x = lim

x→0+ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x

. . . . . .

Example

Findlim

x→0+

√x ln x

limx→0+

√x ln x = lim

x→0+ln x1/

H= lim

x→0+x−1

−12x−3/2

= limx→0+

−2√x = 0

. . . . . .

Indeterminate differences

Example

limx→0+

(1x− cot 2x

This limit is of the form ∞−∞.

SolutionAgain, rig it to make an indeterminate quotient.

limx→0+

sin(2x)− x cos(2x)x sin(2x)

H= lim

x→0+cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)

The limit is +∞ becuase the numerator tends to 1 while thedenominator tends to zero but remains positive.

. . . . . .

Example

limx→0+

(1x− cot 2x

limx→0+

H= lim

. . . . . .

Example

limx→0+

(1x− cot 2x

limx→0+

H= lim

. . . . . .

Example

limx→0+

(1x− cot 2x

limx→0+

H= lim

. . . . . .

Example

limx→0+

(1x− cot 2x

limx→0+

H= lim

. . . . . .

Checking your work

.limx→0

tan 2x2x

= 1, so for small x,

tan 2x ≈ 2x. So cot 2x ≈ 12x

1x− cot 2x ≈ 1

x− 1

→ ∞

as x → 0+.

. . . . . .

Indeterminate powers

Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

limx→0+

(1− 2x)1/x)

= limx→0+

ln((1− 2x)1/x

)= lim

x→0+ln(1− 2x)

This limit is of the form00, so we can use L’Hôpital:

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x1

= −2

This is not the answer, it’s the log of the answer! So the answer wewant is e−2.

. . . . . .

Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

limx→0+

(1− 2x)1/x)

= limx→0+

ln((1− 2x)1/x

)= lim

x→0+ln(1− 2x)

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x1

= −2

. . . . . .

Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

limx→0+

(1− 2x)1/x)

= limx→0+

ln((1− 2x)1/x

)= lim

x→0+ln(1− 2x)

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x1

= −2

. . . . . .

Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

limx→0+

(1− 2x)1/x)

= limx→0+

ln((1− 2x)1/x

)= lim

x→0+ln(1− 2x)

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x1

= −2

This is not the answer, it’s the log of the answer! So the answer wewant is e−2.V63.0121.021, Calculus I (NYU) Section 3.7 L’Hôpital’s Rule November 4, 2010 32 / 35

. . . . . .

Another indeterminate power limit

Example

limx→0

(3x)4x

Solution

ln limx→0+

(3x)4x = limx→0+

ln(3x)4x = limx→0+

4x ln(3x)

= limx→0+

ln(3x)1/4x

H= lim

x→0+

3/3x−1/4x2

= limx→0+

(−4x) = 0

So the answer is e0 = 1.

. . . . . .

Another indeterminate power limit

Example

limx→0

(3x)4x

Solution

ln limx→0+

(3x)4x = limx→0+

ln(3x)4x = limx→0+

4x ln(3x)

= limx→0+

ln(3x)1/4x

H= lim

x→0+

3/3x−1/4x2

= limx→0+

(−4x) = 0

So the answer is e0 = 1.

. . . . . .

Summary

Form Method00 L’Hôpital’s rule directly∞∞ L’Hôpital’s rule directly

0 · ∞ jiggle to make 00 or ∞

∞ .∞−∞ combine to make an indeterminate product or quotient

00 take ln to make an indeterminate product∞0 ditto1∞ ditto

. . . . . .

Final Thoughts

I L’Hôpital’s Rule only works on indeterminate quotientsI Luckily, most indeterminate limits can be transformed into

indeterminate quotientsI L’Hôpital’s Rule gives wrong answers for non-indeterminate limits!

Lesson 17: Indeterminate Forms and L'Hopital's Rule (Section 021 slides)

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