Lesson 25: Indeterminate Forms and L'Hôpital's Rule

Section 4.5Indeterminate Forms and L’Hopital’s Rule

Math 1a

November 26, 2007

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Outline

Indeterminate Forms

L’Hopital’s RuleApplication to Indeterminate ProductsApplication to Indeterminate DifferencesApplication to Indeterminate PowersSummary

The Cauchy Mean Value Theorem (Bonus)

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

Recall






Recall






Recall






We know dividing by zero is bad. Most of the time, if you have anumerator which approaches a finite number and a denominatorwhich approaches zero, the quotient approaches some kind ofinfinity. An exception would be something like

limx→∞

11x sin x

= limx→∞

x sec x .

which doesn’t exist.

Even worse is the situation where the numerator and denominatorboth go to zero.

We know dividing by zero is bad. Most of the time, if you have anumerator which approaches a finite number and a denominatorwhich approaches zero, the quotient approaches some kind ofinfinity. An exception would be something like

limx→∞

11x sin x

= limx→∞

x sec x .

which doesn’t exist.Even worse is the situation where the numerator and denominatorboth go to zero.

Experiments

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.

Experiments

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




Experiments

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




Experiments

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




Experiments

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




Experiments

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




Experiments

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




Experiments

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




Experiments

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




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

I Be careful with the language here. We are not saying that the

limit in each case “is”0

0, and therefore nonexistent because

this expression is undefined.

I The limit is of the form0

0, which means we cannot evaluate it

with our limit laws.

Indeterminate forms are like Tug Of War

Which side wins depends on which side is stronger.

Outline

Indeterminate Forms



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)=

m1

m2=

f ′(x)

g ′(x).

But what if the functions aren’t linear? If only there were a way todeal with functions which were only approximately linear!


limx→a

f (x)

g(x)?


f (x) = m1(x − a)

g(x) = m2(x − a)

Sof (x)

g(x)=

m1

m2=

f ′(x)

g ′(x).



limx→a

f (x)

g(x)?


f (x) = m1(x − a)

g(x) = m2(x − a)

Sof (x)

g(x)=

m1

m2=

f ′(x)

g ′(x).


Theorem (L’Hopital’s Rule)

Suppose f and g are differentiable functions and g ′(x) 6= 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) = ±∞

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 −∞.

L’Hopital’s rule also applies for limits of the form∞∞

.

Theorem (L’Hopital’s Rule)

Suppose f and g are differentiable functions and g ′(x) 6= 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) = ±∞

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 −∞.

L’Hopital’s rule also applies for limits of the form∞∞

.

Meet the Mathematician

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

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

I paid a stipend to JohannBernoulli, who provedthis theorem and namedit after him! Guillaume Franois Antoine,

Marquis de L’Hopital(1661–1704)

How does this affect our examples above?

Example

limx→0

sin2 x

x

H= lim

x→0

2 sin x cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x

(cos x2) (�2x)H= lim

x→0

cos2 x − sin2 x

cos x2 − x2 sin(x2)= 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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x


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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x


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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x

(cos x2) (�2x)

H= lim

x→0

cos2 x − sin2 x


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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x

cos x2 − x2 sin(x2)

= 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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x


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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x


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 cos x

1= 0.

Example

limx→0

sin2 x

sin x2

H= lim

x→0

�2 sin x cos x


x→0

cos2 x − sin2 x


Example

limx→0

sin 3x

sin xH= lim

x→0

3 cos 3x

cos x= 3.

Sketch of Proof of L’Hopital’s Rule

Let x be a number close to a. We know thatf (x)− f (a)

x − a= f ′(c),

for some c ∈ (a, x); alsog(x)− g(a)

x − a= g ′(d), for some

d ∈ (a, x). This means

f (x)

g(x)≈ f ′(c)

g ′(d).

The miracle of the MVT is that a tweaking of it allows us toassume c = d , so that

f (x)

g(x)=

f ′(c)

g ′(c).

The number c depends on x and since it is between a and x , wemust have

limx→a

c(x) = a.

Beware of Red Herrings

Example

Findlimx→0

x

cos x

SolutionThe limit of the denominator is 1, not 0, so L’Hopital’s rule doesnot apply. The limit is 0.

Beware of Red Herrings

Example

Findlimx→0

x

cos x

SolutionThe limit of the denominator is 1, not 0, so L’Hopital’s rule doesnot apply. The limit is 0.

TheoremLet r be any positive number. Then

limx→∞

ex

x r=∞.

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

limx→∞

ex

x r

H= . . .

H= lim

x→∞

ex

r !=∞.

If r is not an integer, let n = [[x ]] and m = n + 1. Then if x > 1,xn < x r < xm, so

ex

xn>

ex

x r>

ex

xm.

Now apply the Squeeze Theorem.


limx→∞

ex

x r=∞.


limx→∞

ex

x r

H= . . .

H= lim

x→∞

ex

r !=∞.


ex

xn>

ex

x r>

ex

xm.



limx→∞

ex

x r=∞.


limx→∞

ex

x r

H= . . .

H= lim

x→∞

ex

r !=∞.


ex

xn>

ex

x r>

ex

xm.


Indeterminate products

Example

Findlim

x→0+

√x ln x

SolutionJury-rig the expression to make an indeterminate quotient. Thenapply L’Hopital’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 products

Example

Findlim

x→0+

√x ln x

SolutionJury-rig the expression to make an indeterminate quotient. Thenapply L’Hopital’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 differences

Example

limx→0

(1

x− cot 2x

)

This limit is of the form ∞−∞, which is indeterminate.

SolutionAgain, rig it to make an indeterminate quotient.

limx→0+

1− x cot 2x

xH= lim

x→0+

2x csc2(2x)− cot(2x)

1= lim

x→0+

2x − cos 2x

sin2 x sin 2xH= lim

x→0+

2 + 2 sin 2x

2 cos 2x sin2 x + 2 cos x sin x sin 2x

=∞


Example

limx→0

(1

x− cot 2x

)This limit is of the form ∞−∞, which is indeterminate.


limx→0+

1− x cot 2x

xH= lim

x→0+


1= lim

x→0+

2x − cos 2x

sin2 x sin 2xH= lim

x→0+

2 + 2 sin 2x


=∞


Example

limx→0

(1

x− cot 2x

)This limit is of the form ∞−∞, which is indeterminate.


limx→0+

1− x cot 2x

xH= lim

x→0+


1= lim

x→0+

2x − cos 2x

sin2 x sin 2xH= lim

x→0+

2 + 2 sin 2x


=∞

Indeterminate powers

Example

limx→0+

(1− 2x)1/x

Take the logarithm:

ln(

limx→0

(1− 2x)1/x)

= limx→0

ln(

(1− 2x)1/x)

= limx→0

1

xln(1− 2x)

This limit is of the form0

0, so we can use L’Hopital:

limx→0

1

xln(1− 2x)

H= lim

x→0

−21−2x

1= −2

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

Indeterminate powers

Example

limx→0+

(1− 2x)1/x

Take the logarithm:

ln(

limx→0

(1− 2x)1/x)

= limx→0

ln(

(1− 2x)1/x)

= limx→0

1

xln(1− 2x)

This limit is of the form0

0, so we can use L’Hopital:

limx→0

1

xln(1− 2x)

H= lim

x→0

−21−2x

1= −2

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

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.

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

Outline

Indeterminate Forms




Apply the MVT to the function

h(x) = (f (b)− f (a))g(x)− (g(b)− g(a))f (x).

We have h(a) = h(b). So there exists a c in (a, b) such thath′(c) = 0. Thus

(f (b)− f (a))g ′(c) = (g(b)− g(a))f ′(c)

This is how L’Hopitalis proved.

Technology

Lesson 25: Indeterminate Forms and L'Hôpital's Rule