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 limx→0h(x)=2. L'Höpital's rule lets us do this the long way, by saying that
limx→0 f(x)/g(x) = limx→0 f'(x)/g'(x) =
limx→0 f''(x)/g''(x) = (limx→0 f''(x)) / (limx→0 g''(x)) =
(limx→0 4) / (limx→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!).
Let f(t),g(t) be functions defined near x that are also differentiable there, and suppose that limt→xf(t) = limt→xg(t) = 0.
limt→x f(t)/g(t) =
limt→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.
A similar version exists when f,g both tend to ∞:
Let f(t),g(t) be functions defined near x that are also differentiable there. Suppose also that limt→xf(t) = limt→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 limt→xG(t)/F(t), which is a limit of the first form.
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 limt→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)
=0 (or f(x)=g(x)=∞)
. The forms "0/0
" and "∞/∞" which then appear are termed "indeterm
inate 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.