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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: $\lim _{x \to \infty} \frac{f(x)}{g(x)}=\lim_{x \to \infty}\frac{f'(x)}{g'(x)}$

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  • $\begingroup$ Have you seen the basic definition of a limit proof of L'Hopital? $\endgroup$ – Andrew Li May 7 '18 at 23:20
  • $\begingroup$ 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. $\endgroup$ – fookyou May 7 '18 at 23:23
  • $\begingroup$ @fookyou Are you looking for the the general or special case? $\endgroup$ – Andrew Li May 7 '18 at 23:28
  • $\begingroup$ i don't know what you call special or general case, i edited my question with the one that i'm concerned with $\endgroup$ – fookyou May 7 '18 at 23:37
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If $f(a) = g(a) = 0$ and $f'(a),$ and $g'(a)$ exist.

$f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a} = \lim_\limits{x\to a} \frac {f(x)}{x-a}\\ g'(a) = \lim_\limits{x\to a} \frac {g(x) - g(a)}{x-a} = \lim_\limits{x\to a} \frac {g(x)}{x-a}\\ \frac {f'(a)}{g'(a)} = \lim_\limits{x\to a} \frac {f(x)}{g(x)}$

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  • $\begingroup$ This is the special case, and the OP seems to be referring to the more general case, maybe we should wait for further clarification? $\endgroup$ – Andrew Li May 7 '18 at 23:29
  • $\begingroup$ thanks for your effort but this is not what i'm looking for unfortunately (and i can't upvote your answer either) $\endgroup$ – fookyou May 7 '18 at 23:38

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