..

Sec on 3.7Indeterminate forms and lHôpital’s

Rule

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

March 30, 2011

Announcements

I Midterm has beenreturned. Please see FAQon Blackboard (under”Exams and Quizzes”)

I Quiz 3 this week inrecita on on Sec on 2.6,2.8, 3.1, 3.2

ObjectivesI Know when a limit is ofindeterminate form:

I indeterminate quo ents: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 quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.

Recall

Recall the limit laws from Chapter 2.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 quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.

Recall

Recall the limit laws from Chapter 2.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 quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.

Recall

Recall the limit laws from Chapter 2.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 quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.

More about dividing limits

I We know dividing by zero is bad.I Most of the me, if an expression’s numerator approaches afinite nonzero number and denominator approaches zero, thequo ent has an infinite. For example:

limx→0+

1x= +∞ lim

x→0−

cos xx3

= −∞

Why 1/0 ̸= ∞Consider the func on f(x) =

11x sin x

.

..x

.

y

Then limx→∞

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

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

Why 1/0 ̸= ∞Consider the func on f(x) =

11x sin x

.

..x

.

y

Then limx→∞

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

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

Experiments with funny limitsI lim

x→0

sin2 xx

= 0

I limx→0

xsin2 x

does not exist

I limx→0

sin2 xsin(x2)

= 1

I limx→0

sin 3xsin x

= 3

.

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

answers in different cases, we say this form is indeterminate.

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

I Be careful with the language here. Weare not saying that the limit in eachcase “is”

00, and therefore nonexistent

because this expression is undefined.

I The limit is of the form00, which means

we cannot evaluate 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

Rela ve Rates of Growth

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

The Linear CaseQues on

If f and g are lines and f(a) = g(a) = 0, what is limx→a

f(x)g(x)

?

Solu onThe func ons f and g can be wri en in the form

f(x) = m1(x− a) g(x) = m2(x− a)

Sof(x)g(x)

=m1

m2

The Linear CaseQues on

If f and g are lines and f(a) = g(a) = 0, what is limx→a

f(x)g(x)

?

Solu onThe func ons f and g can be wri en in the form

f(x) = m1(x− a) g(x) = m2(x− a)

Sof(x)g(x)

=m1

m2

The Linear Case, Illustrated

.. x.

y

.

y = f(x)

.

y = g(x)

..a

..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)

=m1

m2

What then?

I But what if the func ons aren’t linear?

I Can we approximate a func on near a point with a linearfunc on?

I What would be the slope of that linear func on?

Thederiva ve!

What then?

I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?

I What would be the slope of that linear func on?

Thederiva ve!

What then?

I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?

I What would be the slope of that linear func on?

Thederiva ve!

What then?

I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?

I What would be the slope of that linear func on? Thederiva ve!

Theorem of the DayTheorem (L’Hopital’s Rule)

Suppose f and g are differen able func ons and g′(x) ̸= 0 near a(except possibly at a). Suppose that

limx→a

f(x) = 0 and limx→a

g(x) = 0

or 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 military man, butpoor eyesight forced him intomath

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

I paid a s pend to JohannBernoulli, who proved thistheorem and named it a er him!

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

Revisiting the previous examples

Example

limx→0

sin2 xx

H= lim

x→0

2 sin x

..

sin x → 0

cos x1

= 0

Revisiting the previous examples

Example

limx→0

sin2 xx

H= lim

x→0

2 sin x

..

sin x → 0

cos x1

= 0

Revisiting the previous examples

Example

limx→0

sin2 xx

H= lim

x→0

2 sin x..

sin x → 0

cos x1

= 0

Revisiting the previous examples

Example

limx→0

sin2 xx

H= lim

x→0

2 sin x..

sin x → 0

cos x1

= 0

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x..

numerator → 0

sin x2..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x..

numerator → 0

sin x2..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)

H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)

H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x..

numerator → 0

(cos x2) (��2x..

denominator → 0

)

H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x..

numerator → 0

(cos x2) (��2x..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x..

numerator → 1

cos x2 − 2x2 sin(x2)..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin2 x

..

numerator → 0

sin x2

..

denominator → 0

H= lim

x→0

��2 sin x cos x

..

numerator → 0

(cos x2) (��2x

..

denominator → 0

)H= lim

x→0

cos2 x− sin2 x

..

numerator → 1

cos x2 − 2x2 sin(x2)

..

denominator → 1

= 1

Revisiting the previous examples

Example

limx→0

sin 3xsin x

H= lim

x→0

3 cos 3xcos x

= 3.

Revisiting the previous examples

Example

limx→0

sin 3xsin x

H= lim

x→0

3 cos 3xcos x

= 3.

Another Example

Example

Findlimx→0

xcos x

Solu onThe 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

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

Outline

L’Hôpital’s Rule

Rela ve Rates of Growth

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

Limits of Rational Functionsrevisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

106

=53

Limits of Rational Functionsrevisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

106

=53

Limits of Rational Functionsrevisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

106

=53

Limits of Rational Functionsrevisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

106

=53

Limits of Rational Functionsrevisited

Example

Find limx→∞

5x2 + 3x− 13x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 13x2 + 7x+ 27

H= lim

x→∞

10x+ 36x+ 7

H= lim

x→∞

106

=53

Limits of Rational Functionsrevisited II

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

= ∞

Limits of Rational Functionsrevisited II

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

= ∞

Limits of Rational Functionsrevisited II

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

= ∞

Limits of Rational Functionsrevisited II

Example

Find limx→∞

5x2 + 3x− 17x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

5x2 + 3x− 17x+ 27

H= lim

x→∞

10x+ 37

= ∞

Limits of Rational Functionsrevisited III

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

= 0

Limits of Rational Functionsrevisited III

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

= 0

Limits of Rational Functionsrevisited III

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

= 0

Limits of Rational Functionsrevisited III

Example

Find limx→∞

4x+ 73x2 + 7x+ 27

if it exists.

Solu onUsing L’Hôpital:

limx→∞

4x+ 73x2 + 7x+ 27

H= lim

x→∞

46x+ 7

= 0

Limits of Rational FunctionsFactLet 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)

= 0

I If p = q, then limx→∞

f(x)g(x)

is the ra o of the leading coefficients of

f and g.

Exponential vs. geometric growth

Example

Find limx→∞

ex

x2, if it exists.

Solu onWe have

limx→∞

ex

x2H= lim

x→∞

ex

2xH= lim

x→∞

ex

2= ∞.

Exponential vs. geometric growth

Example

Find limx→∞

ex

x2, if it exists.

Solu onWe have

limx→∞

ex

x2H= lim

x→∞

ex

2xH= lim

x→∞

ex

2= ∞.

Exponential vs. geometric growthExample

What about limx→∞

ex

x3?

AnswerS ll∞. (Why?)

Solu on

limx→∞

ex

x3H= lim

x→∞

ex

3x2H= lim

x→∞

ex

6xH= lim

x→∞

ex

6= ∞.

Exponential vs. geometric growthExample

What about limx→∞

ex

x3?

AnswerS ll∞. (Why?)

Solu on

limx→∞

ex

x3H= lim

x→∞

ex

3x2H= lim

x→∞

ex

6xH= lim

x→∞

ex

6= ∞.

Exponential vs. geometric growthExample

What about limx→∞

ex

x3?

AnswerS ll∞. (Why?)

Solu on

limx→∞

ex

x3H= lim

x→∞

ex

3x2H= lim

x→∞

ex

6xH= lim

x→∞

ex

6= ∞.

Exponential vs. fractional powersExample

Find limx→∞

ex√x, if it exists.

Exponential vs. fractional powersExample

Find limx→∞

ex√x, if it exists.

Solu on (without L’Hôpital)

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

ex

x1/2>

ex

x

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

Exponential vs. fractional powersExample

Find limx→∞

ex√x, if it exists.

Solu on (with L’Hôpital)

limx→∞

ex√x= lim

x→∞

ex12x−1/2 = lim

x→∞2√xex = ∞

Exponential vs. any powerTheorem

Let r be any posi ve number. Then limx→∞

ex

xr= ∞.

Proof.

Exponential vs. any powerTheorem

Let r be any posi ve number. Then limx→∞

ex

xr= ∞.

Proof.If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac-on. You get

limx→∞

ex

xrH= . . .

H= lim

x→∞

ex

r!= ∞.

Exponential vs. any powerTheorem

Let r be any posi ve number. Then limx→∞

ex

xr= ∞.

Proof.If r is not an integer, letm be the smallest integer greater than r. Thenif x > 1, xr < xm, so

ex

xr>

ex

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

previous step.

Any exponential vs. any powerTheorem

Let a > 1 and r > 0. Then limx→∞

ax

xr= ∞.

Proof.If r is a posi ve integer, we have

limx→∞

ax

xrH= . . .

H= lim

x→∞

(ln a)rax

r!= ∞.

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

So even limx→∞

(1.00000001)x

x100000000= ∞!

Any exponential vs. any powerTheorem

Let a > 1 and r > 0. Then limx→∞

ax

xr= ∞.

Proof.If r is a posi ve integer, we have

limx→∞

ax

xrH= . . .

H= lim

x→∞

(ln a)rax

r!= ∞.

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

So even limx→∞

(1.00000001)x

x100000000= ∞!

Any exponential vs. any powerTheorem

Let a > 1 and r > 0. Then limx→∞

ax

xr= ∞.

Proof.If r is a posi ve integer, we have

limx→∞

ax

xrH= . . .

H= lim

x→∞

(ln a)rax

r!= ∞.

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

So even limx→∞

(1.00000001)x

x100000000= ∞!

Logarithmic versus power growthTheorem

Let r be any posi ve number. Then limx→∞

ln xxr

= 0.

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

limx→∞

ln xxr

H= lim

x→∞

1/xrxr−1 = lim

x→∞

1rxr

= 0.

Logarithmic versus power growthTheorem

Let r be any posi ve number. Then limx→∞

ln xxr

= 0.

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

limx→∞

ln xxr

H= lim

x→∞

1/xrxr−1 = lim

x→∞

1rxr

= 0.

Outline

L’Hôpital’s Rule

Rela ve Rates of Growth

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x

= limx→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2 = lim

x→0+−2

√x = 0

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x

= limx→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2 = lim

x→0+−2

√x = 0

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x = lim

x→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2 = lim

x→0+−2

√x = 0

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x = lim

x→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2

= limx→0+

−2√x = 0

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x = lim

x→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2 = lim

x→0+−2

√x

= 0

Indeterminate productsExample

Find limx→0+

√x ln x

This limit is of the form 0 · (−∞).

Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:

limx→0+

√x ln x = lim

x→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2 = lim

x→0+−2

√x = 0

Indeterminate differences

Example

limx→0+

(1x− cot 2x

)This limit is of the form∞−∞.

Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.

limx→0+

(1x− cot 2x

)= lim

x→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 +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.

Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.

limx→0+

(1x− cot 2x

)= lim

x→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 +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.

Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.

limx→0+

(1x− cot 2x

)= lim

x→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 +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.

Indeterminate DifferencesSolu onAgain, rig it to make an indeterminate quo ent.

limx→0+

(1x− cot 2x

)= lim

x→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 +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.

Checking your workThis all goes in the thought cloud

limx→0

tan 2x2x

= 1, so for small x, tan 2x ≈ 2x. So cot 2x ≈ 12x

and

1x− cot 2x ≈ 1

x− 1

2x=

12x

→ ∞

as x → 0+.

Indeterminate powersExample

Find limx→0+

(1− 2x)1/x

Solu on

Indeterminate powersExample

Find limx→0+

(1− 2x)1/x

Solu onTake the logarithm:

ln(

limx→0+

(1− 2x)1/x)

= limx→0+

ln((1− 2x)1/x

)= lim

x→0+

ln(1− 2x)x

Indeterminate powersExample

Find limx→0+

(1− 2x)1/x

Solu on

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

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x

1= −2

Indeterminate powersExample

Find limx→0+

(1− 2x)1/x

Solu on

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

limx→0+

ln(1− 2x)x

H= lim

x→0+

−21−2x

1= −2

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

Another indeterminate power limitExample

Find limx→0+

(3x)4x

Solu on

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 limitExample

Find limx→0+

(3x)4x

Solu on

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.

SummaryForm Method

00 L’Hôpital’s rule directly∞∞ L’Hôpital’s rule directly

0 · ∞ jiggle to make 00 or

∞∞ .

∞−∞ combine to make an indeterminate product or quo ent

00 take ln to make an indeterminate product

∞0 di o

1∞ di o

Final Thoughts

I L’Hôpital’s Rule only works on indeterminate quo entsI Luckily, most indeterminate limits can be transformed intoindeterminate quo ents

I L’Hôpital’s Rule gives wrong answers for non-indeterminatelimits!