l'Hôpital's rule

Applied when one or more indeterminate forms such as 0/0 or INFINITY/INFINITY are encountered. Suppose lim(x->0) f(x)/g(x)=0/0, then
simply take lim(x->0) f'(x)/g'(x). Repeat until an indeterminate form is not reached.

This is actually a sort of misnamed rule. In 1694, John Bernoulli decided to work for his former student l'Hopital to solve problems for him and keep him generally up to date on various advancements in Calculus. One of the problems that l'Hopital gave him was the indeterminate form problem, which Bernoulli solved as promised. Later, l'Hopital published his notes on calculus in a book and the rule for solving limits with indeterminate forms was part of it. Now, l'Hopital wasn't such a bad guy, he did acknowledge his debt to Bernoulli and even published the book anonymously so he wouldn't claim ownership. Despite this, Bernoulli went nuts, accusing l'Hopital of plagiarism. Unfortunately, this accusation was inadvertently supported after l'Hopital's death in 1704 by the publisher's promotion of the book as l'Hopital's. Quite famous as a jealous man, Bernoulli insisted that he was the author of the entire work. History accepted Bernoulli's claim (until recently), but still named the rule after l'Hopital.

L'Höpital's rule is the name given to some rather easy theorems. The mystique stems from the fact that they appear to allow you to compute "0/0" or "∞/∞". Of course, they do nothing of the sort!

Suppose we have f(x)=2x², g(x)=x². Then h(x)=f(x)/g(x)=2, except when x=0, where it is undefined. In the spirit of analytic continuation, we would like to remove this "blemish" from h, by saying that h(0)=2, except that (of course) it doesn't. Instead, we shall content ourselves with showing that lim_x→0h(x)=2. L'Höpital's rule lets us do this the long way, by saying that

lim_x→0 f(x)/g(x) = lim_x→0 f'(x)/g'(x) =
lim_x→0 f''(x)/g''(x) = (lim_x→0 f''(x)) / (lim_x→0 g''(x)) =
(lim_x→0 4) / (lim_x→0 2) = 2,

where the existence of each limit follows from the existence of the limit to its right (i.e. we even write the calculation backwards!).

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1. Theorem. Let f(t),g(t) be functions defined near x that are also differentiable there, and suppose that lim_t→xf(t) = lim_t→xg(t) = 0. Then

   lim_t→x f(t)/g(t) = lim_t→x f'(t)/g'(t).    (*)

   (In particular, the first limit exists iff the second limit does). In particular, if the k'th derivatives f^(k) and g^(k) are continuous and non-zero at x then we may compute the limit by a simple substitution.
2. A similar version exists when f,g both tend to ∞:
   Theorem. Let f(t),g(t) be functions defined near x that are also differentiable there. Suppose also that lim_t→xf(t) = lim_t→xg(t)=∞. Then (*) holds, just as in the previous case.

   The proof of this form follows immediately from the previous form, taking F(t)=1/f(t) and G(t)=1/g(t); the limit to be calculated in this case is precisely lim_t→xG(t)/F(t), which is a limit of the first form.

3. FINALLY, forms also exist for the limit at ∞ (i.e. writing "∞" instead of "x" in the preceding). These are also consequences of the first form, this time defining F(t)=f(1/t) and G(t)=g(1/t), and using the first form for F(t)/G(t) near x=0.

IMPORTANT NOTE: (All) the conditions are important!

In fact l'Hôpital's rule is less useful than might be expected! It only works in the "easy" cases.

Probably the best-known "example" is to compute lim_t→0 sin(t)/t = 1. While taking the derivatives and substituting seems to work, in actual fact this is precisely the limit that must be computed for proving sin'(t) = cos(t). So using l'Hôpital's rule here is just circular reasoning...

Sometimes people write the nonsense "f(x)/g(x)", and claim that

f(x)/g(x) = f'(x)/g'(x)

when f(x)=g(x)=0 (or f(x)=g(x)=∞). The forms "0/0" and "∞/∞" which then appear are termed "indeterminate forms". But this is meaningless.

There is no consistent, meaningful sense in which we can interpret an expression such as "0/0". Indeed, if there is any lesson to be learnt from l'Hôpital's rule, it is that even with calculus, the value which we would wish to assign to "0/0" depends precisely on f(t) and g(t) above. But neither f(t) nor g(t) appear when writing "0/0"!. So "0/0" differs substantially from "1+1". The difference stems from the fact that the second has meaning. I cannot divide by zero. Sorry.