On-line Math 21

4.4  l'Hôpital's Rule

This is a nice, computational formula that can be used to compute limits, and so is often taught in high-school calculus classes. It usually isn't justified there, however. It also isn't usually mentioned how the name came about. In the 17^th and 18^th centuries, it was common for kings and other rulers, rather than universities, to employ mathematicians. l'Hôpital was one such employer, who hired a mathematician by the name of Johann Bernoulli to prove him a theorem. Bernoulli did, and the result was named for the nobleman who commissioned the work rather than the mathematician who thought it up. In fairness, l'Hôpital was a mathematician in his own right. He wrote the first definitive text in differential calculus (but he is not to blame for the existence of calculus courses). On the other hand, he did hire Bernoulli to teach him calculus, and to provide him with results which he published in his text as his own.

l'Hôpital's Rule is used to make sense out of limits that are indeterminate, that is, limits that you can't figure out by just plugging in the limiting value of x into the expression. We have seen many such limits, usually of the form

0 0 ,

that is, a limit:

lim x® a  f(x) g(x)

where both

lim x® a  f(x) = 0

and

lim x® a  g(x) = 0.

a here could be a number, or ±¥. It's important to make the distinction between making sense of a limit of the form 0/0 versus actually making sense of the fraction itself. 0/0 is only understandable as a limit.

A limit is indeterminate if it is one of the following forms (that is, the pieces of the expression have the indicated limits themselves):

0 0 ,   ¥ ¥ ,  1¥,  ¥-¥,  0×¥,  00,  0¥,  or  ¥0.

By some standard algebraic tricks, any of the other forms can be reduced to one of the first two (though the last two should be kept separate).

Here is what l'Hôpital paid for:

Theorem 1 Let f and g be differentiable on some interval, with a in that interval, and assume they satisfy

lim x® a  f(x) = 0,  and   lim x® a  g(x) = 0.

Then,

lim x® a  f(x) g(x) = lim x® a  f¢(x) g¢(x) .

Remark 1 a could be a number, or it could be ¥. The theorem also holds for one-sided limits, like

lim x® a+  f(x) g(x) ,

as well.

Proof.

Example 1

lim x® 0  sin(3x) tan(4x)

Solution

Exercise 1

lim x® 2  Öx-2 x2-3x+2 =

Exercise 2

lim x® 0  cos(x)-1 x2 =

Hint:

Here, once you apply l'Hôpital's rule, you will still be left with an indeterminate limit. Apply l'Hôpital's rule again. You can continue to do this until you come to a fraction you can simplify, or until the limit is no longer an indeterminate form.

6.0.1  Infinity/infinity

There is a separate l'Hôpital's rule theorem for for ¥/¥ -type limits,

lim x® a  f(x) g(x) , where  lim x® a  f(x) = ¥ = lim x® a  g(x).

But the statement is essentially the same. The proof is a bit more subtle; it can't quite be solved by turning it into the limit

lim x® a  1/g(x) 1/f(x) ,

since that trick requires that you know that the limit exists in order to show how to compute it. l'Hôpital's rule actually does not require in advance that you know the limit exists. It says that the limit exists and equals the limit of the quotient of the derivatives, if that latter limit exists.

Theorem 2 [l'Hôpital's Rule, ¥/¥ form] Let f and g be differentiable on some interval, with a in that interval, and assume they satisfy

lim x® a  f(x) = ¥,  and   lim x® a  g(x) = ¥.

Then,

lim x® a  f(x) g(x) = lim x® a  f¢(x) g¢(x) .

Example 2

lim x® ¥  x3 ex

Solution

Example 3

lim x® ¥  ln(x) x2

Solution

Example 4

lim x® ¥  æ è Ö   x2+x+1   - Ö   x2-x   ö ø

Solution

Example 5

lim x® ¥  æ ç è 1+ 1 x ö ÷ ø x

Solution

Exercise 3

lim x® ¥  æ ç è 1+ r x ö ÷ ø x   =

7.0.2  Growth rates of functions.

For functions that go to infinity as x® ¥, we can use l'Hôpital's rule to compare how quickly they go to infinity. We say that f << g (or, g >> f ) if

lim x® ¥  f(x) g(x) = 0,

or, equivalently,

lim x® ¥  g(x) f(x) = ¥.

As an example, since

lim x® ¥  ex x2 = lim x® ¥  ex 2x = lim x® ¥  ex 2 = ¥,

we say that e^x >> x² , as you might expect.

Exercise 4 Place the following functions in order from the slowest growth rate to the fastest, as x® ¥.

ex, x2, ln(x), xx, xln(x), exln(x).

You do not have to use l'Hôpital's rule in every case.

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