proof of l'Hôpital's rule [duplicate]

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.

Solution 1:

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 $\epsilon = 1/M$).