Guillaume De l'Hôpital1661 - 1704

8.2L’Hôpital’s Rule

Actually, L’Hôpital’s Rule was developed by his teacher Johann Bernoulli. De l’Hôpital paid Bernoulli for private lessons, and then published the first Calculus book based on those lessons.

Zero divided by zero can not be evaluated, and is an example of indeterminate form.

2

2

4lim

2x

x

x

Consider:

If we try to evaluate this by direct substitution, we get:0

0

In this case, we can evaluate this limit by factoring and canceling:

2

2

4lim

2x

x

x

2

2 2lim

2x

x x

x

2lim 2x

x

4

2

2

4lim

2x

x

x

limx a

f x

g x

2

2

4lim

2x

dx

dxdx

dx

2

2lim

1x

x

4

L’Hôpital’s Rule:

If is indeterminate, then:

limx a

f x

g x

lim limx a x a

f x f x

g x g x

We can confirm L’Hôpital’s rule by working backwards, and using the definition of derivative:

f a

g a

lim

lim

x a

x a

f x f a

x ag x g a

x a

limx a

f x f a

x ag x g a

x a

limx a

f x f a

g x g a

0lim

0x a

f x

g x

limx a

f x

g x

Example:

20

1 coslimx

x

x x

0

sinlim

1 2x

x

x

0

If it’s no longer indeterminate, then STOP!

If we try to continue with L’Hôpital’s rule:

0

sinlim

1 2x

x

x

0

coslim

2x

x

1

2 which is wrong,

wrong, wrong!

On the other hand, you can apply L’Hôpital’s rule as many times as necessary as long as the fraction is still indeterminate:

20

1 12lim

x

xx

x

1

2

0

1 11

2 2lim2x

x

x

0

0

0

0

0

0not

1

2

20

11 1

2limx

x x

x

3

2

0

11

4lim2x

x

14

2

1

8

(Rewritten in exponential form.)

L’Hôpital’s rule can be used to evaluate other indeterminate0

0forms besides .

What makes an expression indeterminate?

lim1000x

x

Consider:

We can hold one part of the expression constant:

1000lim 0x x

There are conflicting trends here. The actual limit will depend on the rates at which the numerator and denominator approach infinity, so we say that an expression in this form is indeterminate.

Let’s look at another one:

0

0lim1000 1x

x

Consider:

We can hold one part of the expression constant:

0.1limxx

Once again, we have conflicting trends, so this form is indeterminate.

0.1lim 0xx

Finally, here is an expression that looks like it might be indeterminate :

0

lim .1 0x

x

Consider:

We can hold one part of the expression constant:

lim .1 0x

x

The limit is zero any way you look at it, so the expression is not indeterminate.

1000

0lim 0xx

Here is the standard list of the seven indeterminate forms:

0 1 00 0

0

0

There are other indeterminate forms using complex numbers, but those are beyond the scope of this class.

These expressions have to be changed to a fraction before using L’Hopital’s Rule!!!

These expressions can be evaluated “as is” using L’Hopital’s Rule.

1lim sinx

xx

This approaches0

0

1sin

lim1x

x

x

This approaches 0

We already know that0

sinlim 1x

x

x

but if we want to use L’Hôpital’s rule:

2

2

1 1cos

lim1x

x x

x

1sin

lim1x

x

x

1lim cosx x

cos 0 1

1

1 1lim

ln 1x x x

If we find a common denominator and subtract, we get:

1

1 lnlim

1 lnx

x x

x x

Now it is in the form0

0

This is indeterminate form

1

11

lim1

lnx

xx

xx

L’Hôpital’s rule applied once.

0

0Fractions cleared. Still

1

1lim

1 lnx

x

x x x

1

1 1lim

ln 1x x x

1

1 lnlim

1 lnx

x x

x x

1

11

lim1

lnx

xx

xx

1

1lim

1 1 lnx x

L’Hôpital again.

1

2

1

1lim

1 lnx

x

x x x

Indeterminate Forms: 1 00 0

Evaluating these forms requires a mathematical trick to change the expression into a fraction.

ln lnnu n u

When we take the log of an exponential function, the exponent can be moved out front.

ln1u

n

We can then write the expression as a fraction, which allows us to use L’Hôpital’s rule.

limx a

f x

ln limx a

f xe

lim lnx a

f xe

We can take the log of the function as long as we exponentiate at the same time.

Then move the limit notation outside of the log.

Indeterminate Forms: 1 00 0

1/lim x

xx

1/lim ln x

xx

e

1lim lnx

xxe

lnlimx

x

xe

1

lim1x

x

e

0e

1

0

L’Hôpitalapplied

Example: