Lesson 17: Indeterminate Forms and L'Hôpital's Rule

Section 3.7Indeterminate Forms and L’Hopital’s

Rule

V63.0121.002.2010Su, Calculus I

New York University

June 7, 2010

Announcements

I

Announcements

I

V63.0121.002.2010Su, Calculus I (NYU) L’Hopital’s Rule June 7, 2010 2 / 26

Objectives

I Know when a limit is ofindeterminate form:

I indeterminate quotients:0/0, ∞/∞

I indeterminate products:0×∞

I indeterminate differences:∞−∞

I indeterminate powers: 00,∞0, and 1∞


Experiments with funny limits

I limx→0

sin2 x

x

= 0

I limx→0

x

sin2 x

does not exist

I limx→0

sin2 x

sin(x2)

= 1

I limx→0

sin 3x

sin x

= 3

All of these are of the form0

0, and since we can get different answers in

different cases, we say this form is indeterminate.



I limx→0

sin2 x

x= 0

I limx→0

x

sin2 x

does not exist

I limx→0

sin2 x

sin(x2)

= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x

sin2 x

does not exist

I limx→0

sin2 x

sin(x2)

= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x

sin2 xdoes not exist

I limx→0

sin2 x

sin(x2)

= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x


I limx→0

sin2 x

sin(x2)

= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x


I limx→0

sin2 x

sin(x2)= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x


I limx→0

sin2 x

sin(x2)= 1

I limx→0

sin 3x

sin x

= 3






I limx→0

sin2 x

x= 0

I limx→0

x


I limx→0

sin2 x

sin(x2)= 1

I limx→0

sin 3x

sin x= 3






I limx→0

sin2 x

x= 0

I limx→0

x


I limx→0

sin2 x

sin(x2)= 1

I limx→0

sin 3x

sin x= 3





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 limits

I 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







Recall







Recall







More about dividing limits

I We know dividing by zero is bad.

I Most of the time, if an expression’s numerator approaches a finitenumber and denominator approaches zero, the quotient approachessome kind of infinity. For example:

limx→0+

1

x= +∞ lim

x→0−

cos x

x3= −∞

I An exception would be something like

limx→∞

11x sin x

= limx→∞

x csc x .

which doesn’t exist.

I Even less predictable: numerator and denominator both go to zero.





limx→0+

1

x= +∞ lim

x→0−

cos x

x3= −∞


limx→∞

11x sin x

= limx→∞

x csc x .







limx→0+

1

x= +∞ lim

x→0−

cos x

x3= −∞


limx→∞

11x sin x

= limx→∞

x csc x .




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

I Be careful with the languagehere. We are not saying thatthe limit in each case “is”0

0, and therefore nonexistent

because this expression isundefined.

I The limit is of the form0

0,

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’Hopital’s Rule

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

Summary


The Linear Case

Question

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

limx→a

f (x)

g(x)?

Solution

The functions f and g can be written in the form

f (x) = m1(x − a)

g(x) = m2(x − a)

Sof (x)

g(x)=

m1

m2


The Linear Case

Question

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

limx→a

f (x)

g(x)?

Solution

The functions f and g can be written 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 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?





What then?



I What would be the slope of that linear function?

The derivative!


What then?





Theorem of the Day

Theorem (L’Hopital’s Rule)

Suppose f and g are differentiable functions and g ′(x) 6= 0 near a (exceptpossibly 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) = ±∞

Then

limx→a

f (x)

g(x)= lim

x→a

f ′(x)

g ′(x),

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


Meet the Mathematician: L’Hopital

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 Francois Antoine,

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


Revisiting the previous examples

Example

limx→0

sin2 x

x

H= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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

Example

limx→0

sin 3x

sin x

H= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1

= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1

= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.



Example

limx→0

sin2 x

xH= lim

x→0

2 sin x

sin x → 0

cos x

1= 0

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


denominator→ 1

= 1

Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.


Another Example

Example

Findlimx→0

x

cos x

Solution

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


Beware of Red Herrings

Example

Findlimx→0

x

cos x

Solution

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


Outline

L’Hopital’s Rule

Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers

Summary


Indeterminate products

Example

Findlim

x→0+

√x ln x

This limit is of the form 0 · (−∞).

Solution

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

limx→0+

√x ln x

= limx→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2

= limx→0+

−2√

x = 0



Example

Findlim

x→0+

√x ln x


Solution


limx→0+

√x ln x

= limx→0+

ln x1/√x

H= lim

x→0+

x−1

−12x−3/2

= limx→0+

−2√

x = 0



Example

Findlim

x→0+

√x ln x


Solution


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



Example

Findlim

x→0+

√x ln x


Solution


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



Example

Findlim

x→0+

√x ln x


Solution


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



Example

Findlim

x→0+

√x ln x


Solution


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 differences

Example

limx→0+

(1

x− cot 2x

)

This limit is of the form ∞−∞.

Solution

Again, 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 the denominatortends to zero but remains positive.



Example

limx→0+

(1

x− cot 2x

)


Solution


limx→0+


x sin(2x)

H= lim

x→0+



=∞




Example

limx→0+

(1

x− cot 2x

)


Solution


limx→0+


x sin(2x)H= lim

x→0+



=∞




Example

limx→0+

(1

x− cot 2x

)


Solution


limx→0+


x sin(2x)H= lim

x→0+



=∞




Example

limx→0+

(1

x− cot 2x

)


Solution


limx→0+


x sin(2x)H= lim

x→0+



=∞



Checking your work

limx→0

tan 2x

2x= 1, so for small x ,

tan 2x ≈ 2x . So cot 2x ≈ 1

2xand

1

x− cot 2x ≈ 1

x− 1

2x=

1

2x→∞

as x → 0+.


Indeterminate powers

Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

ln

(lim

x→0+(1− 2x)1/x

)= lim

x→0+ln(

(1− 2x)1/x)

= limx→0+

ln(1− 2x)

x

This limit is of the form0

0, so we can use L’Hopital:

limx→0+

ln(1− 2x)

xH= lim

x→0+

−21−2x

1= −2

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



Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

ln

(lim

x→0+(1− 2x)1/x

)= lim

x→0+ln(

(1− 2x)1/x)

= limx→0+

ln(1− 2x)

x



limx→0+

ln(1− 2x)

xH= lim

x→0+

−21−2x

1= −2




Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

ln

(lim

x→0+(1− 2x)1/x

)= lim

x→0+ln(

(1− 2x)1/x)

= limx→0+

ln(1− 2x)

x



limx→0+

ln(1− 2x)

xH= lim

x→0+

−21−2x

1= −2




Example

Find limx→0+

(1− 2x)1/x

Take the logarithm:

ln

(lim

x→0+(1− 2x)1/x

)= lim

x→0+ln(

(1− 2x)1/x)

= limx→0+

ln(1− 2x)

x



limx→0+

ln(1− 2x)

xH= lim

x→0+

−21−2x

1= −2



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 Method

00 L’Hopital’s rule directly

∞∞ L’Hopital’s rule directly

0 · ∞ jiggle to make 00 or ∞∞ .

∞−∞ factor to make an indeterminate product

00 take ln to make an indeterminate product

∞0 ditto

1∞ ditto


Final Thoughts

I L’Hopital’s Rule only works on indeterminate quotients

I Luckily, most indeterminate limits can be transformed intoindeterminate quotients

I L’Hopital’s Rule gives wrong answers for non-indeterminate limits!


Technology

Lesson 17: Indeterminate Forms and L'Hôpital's Rule