Guillaume De l'Hôpital1661 - 1704

8.2 L’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

If we zoom in far enough, the curves will appear as straight lines.

2

2

4lim

2x

x

x

The limit is the ratio of the numerator over the denominator as x approaches 2.

2 4x

2x

limx a

f x

g x

Notice the ratio of the y-coords. of f and g is 0/0 graphically as well!

2

2

4lim

2x

x

x

limx a

f x

g x

f x

g x f x

g x

As 2x

becomes:

2

2

4lim

2x

x

x

limx a

f x

g x

As 2x

f x

g xbecomes:

df

dg

df

dg

dx

dxdf

xdgd

As we zoom in super close…f becomes a small change in f with respect to the x-axis. Same goes for g….notice the ratio of dx remains the same.

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

The only forms that work with L’Hopital 0 or

0

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

Ummmm, cool Mr. G…why are we learning this?

Check out the following problem from our past:

0

sinlimx

x

x

Please graph on your GC and try to recall what theorem we used to show the limit exists.

Let’s use our new found friend to find the limit!

Find the following limits…be careful to only apply L’Hopital where direct substitution leads to an indeterminate form.

3

0

1lim5

x

x

ex 3

35 2lim7 1xxx

5

30

8 12lim3x

xx x

2

lnlimx

xx

Try this one out!!!

This will serve you well on the AP and improper integrals (forthcoming) so that you do not even need to spend time visualizing the behavior of some heinous function….but be careful not to use this on any old limit…only on indeterminate forms!!!

20

1 12lim

x

xx

x

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

0forms besides .

The following are also considered indeterminate:

0 1 00 0

The first one, , can be evaluated just like .

0

0

The others must be changed to fractions first.

1lim tanx

xx

This approaches0

0

1tan

lim1x

x

x

This approaches 0

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

Indeterminate Forms: 1 00 0

Evaluating these forms requires a mathematical trick to change the expression into a fraction. We want to bring the exponents “out from the clouds”…what function does this for us?

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:

Try one more!

4

4lim 3 4

x

xx

Add #’s 41,44,47 to hw!!!