# proof of l'Hôpital's rule

I've never worked out the general proof of l'Hôpital before, and am taking the opportunity to do so before I return to the soul-crushing grind of college life.

I looked in Spivak and found a fine proof of the $(x \rightarrow a$ and $f(x), g(x) \rightarrow 0)$ case; but when he went to prove the $(x \rightarrow \infty, f(x), g(x) \rightarrow 0)$ case, he made a suggestion which actually does not work. (He proposed applying l'Hôpital for the first case above to $f(1/x) / g(1/x)$ as $x \rightarrow 0$, but since he used the fact that $f(0)$ and $g(0)$ could be defined to prove the first case, his version doesn't seem to apply. Correct me if I'm wrong!)

I adapted this proof of the second case.

Theorem: Suppose that $\lim_{x \to \infty} f(x), \lim_{x \to \infty} g(x) = 0$, and $\lim_{x \to \infty} \frac{f'(x)}{g'(x)} = l$. Then $\lim_{x \to \infty} \frac{f(x)}{g(x)} = l.$

Proof: Note as a lemma that there exists an interval $I = (c, \infty)$ on which

1. $g'(x) \neq 0$
2. $g(x) \neq 0$

(The hypothesis implies (1); to prove (2), note that every interval $(\eta, \infty)$ would otherwise have $g(\xi) = 0$, so taking any $\eta ' > \eta$, there would be $g'(x) = 0$ on $(\xi, \xi ')$, and thus on $(\eta, \infty)$, against (1).)

Fix $\epsilon > 0$. By hypothesis, $\exists M: x > M \implies |\frac{f'(x)}{g'(x)} - l| < \frac{\epsilon}{2}$. If $M < c$, choose $M = c$ so $M \in I$.

Now pick any $y > M$. By Cauchy, $\forall x > y$ we get $($for $\mu \in (y, x))$

$$\frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\mu)}{g'(\mu)}.$$

But if we take $\lim_{x \to \infty}$ of both sides, we find

$$\lim_{x \to \infty}\frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f(y)}{g(y)} = \lim_{x \to \infty} \frac{f'(\mu)}{g'(\mu)}.$$

We know that the RHS limit is determinate; we also know that

$$l - \frac{\epsilon}{2} < \frac{f'(\mu)}{g'(\mu)} < l + \frac{\epsilon}{2} \implies l - \frac{\epsilon}{2} \leq \lim_{x \to \infty} \frac{f'(\mu)}{g'(\mu)} \leq l + \frac{\epsilon}{2}$$

$$\therefore l - \epsilon < \frac{f(y)}{g(y)} < l + \epsilon .$$

We have therefore furnished $M : |\frac{f(x)}{g(x)} - l| < \epsilon$ if $x > M$. This proves the theorem.

I now feel a sudden urge to begin overusing l'Hôpital! Someone, please tell me if this is correct, so I can go at it.

-
+1, if only for "soul-crushing grind of college life". I sincerely hope you were being facetious (else I'd change my upvote to a downvote). –  Rick Decker Aug 8 '12 at 0:54
@Rick I thought that would be a winner :D –  Chris Aug 8 '12 at 2:04
I upvoted since you bothered to check l'Hôpital; not too many students do that. –  Ｊ. Ｍ. Aug 8 '12 at 3:17

I don't see any major problems with your proof. Though, expanding on Spivak's hint, $f(1/x)g(1/x)$ doesn't work immediately. But, note that $$\lim_{x \to 0^+} f(1/x) = \lim_{u \to \infty} f(u) = 0$$ and similarly for $g$. Hence the functions $$f^*(x) = \lim_{z\to x} f(1/z),$$ $$g^*(x) = \lim_{z\to x} g(1/z)$$ do satisfy the hypotheses for L'Hopital on some interval $[0,\epsilon]$ for $\epsilon > 0$, (think of taking the reciprocal of the interval $[M,\infty]$, then $\epsilon = 1/M$).
Well, it seems that you're suggesting we patch the continuity hypothesis by taking advantage of the differentiability of $f$ for real arguments which are large (considering the reciprocals of small numbers), and then defining $f^*(0) = \lim_{z \to x}f(\frac{1}{z})$. Am I correct on that? –  Chris Aug 8 '12 at 2:28
Yes, (assuming you mean $f^*(0) = \lim_{z\to 0} f(1/z)$ of course). –  nullUser Aug 8 '12 at 2:29