# Proof of De L'hopitals rule which doesn't use the Cauchy MVT or Rolles Theorem

I'm trying to find a proof of De L'Hospital Theorem which doesn't use either the Cauchy MVT or Rolles Theorem.

Does such a thing exist? I've seen some "proofs" using Taylor expansions however they seem to be wrong since Taylor expansions require continuous differentiability....

What needs to be shown is that:

Let $\ a\in \mathbb{R}$ and let the functions $f,g:(a, \infty) \to \mathbb{R}$ on the open interval $(a,\infty)$ continous and differentiable. It holds that $\lim _{x\to \infty} g(x) = +\infty$ and the limit $\lim _{x \to \infty} \frac{f'(x)}{g'(x)}=L$ exists

Then it is to show that: $\lim _{x \to \infty} \frac{f(x)}{g(x)}=\lim_{x \to \infty}\frac{f'(x)}{g'(x)}$

• Have you seen the basic definition of a limit proof of L'Hopital? – Andrew Li May 7 '18 at 23:20
• Have you read my question? There are no limit proofs of L'Hopital which don't use the MVT or Rolles Theorem. So I'm asking if anybody knows one. – fookyou May 7 '18 at 23:23
• @fookyou Are you looking for the the general or special case? – Andrew Li May 7 '18 at 23:28
• i don't know what you call special or general case, i edited my question with the one that i'm concerned with – fookyou May 7 '18 at 23:37

If $f(a) = g(a) = 0$ and $f'(a),$ and $g'(a)$ exist.
$f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a} = \lim_\limits{x\to a} \frac {f(x)}{x-a}\\ g'(a) = \lim_\limits{x\to a} \frac {g(x) - g(a)}{x-a} = \lim_\limits{x\to a} \frac {g(x)}{x-a}\\ \frac {f'(a)}{g'(a)} = \lim_\limits{x\to a} \frac {f(x)}{g(x)}$