Rearding L'Hopital's rule for the $\frac{\infty}{\infty}$ case: Assume $a<c<b$ and $f,g$ are differentiable functions on $(a,b)$\{$c$} such that $g'(x)\neq 0$ for all $x$. If $\lim_{x\to c}g(x)=\infty$ and there is an $L$ in $\mathbb{R}$ such that $$\lim_{x \to c}\frac{f'(x)}{g'(x)}=L$$ implies $$\lim_{x\to c}\frac{f(x)}{g(x)}=L$$ Assuming $L=\infty$, does this remain valid? Are there any counterexamples? I can't think of any so am going to start working on a proof that it remains valid in this case but wondering if I'm missing an obvious exception.
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You mean $\displaystyle \lim_{x \to c} \dfrac{f'(x)}{g'(x)} = L$? – azif00 Apr 13 '20 at 02:26
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yes. fixed. thank you. – mayalarson Apr 13 '20 at 02:39
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If $L=\infty,$ then I must change this way. I think that we can have some other types: $\dfrac{0}{0}, 1^{\infty}, 0^0,\infty^0,\infty-\infty.$ You can reference in link. – Tran Nam Son Apr 13 '20 at 02:52
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Usually, when one writes things like $\lim_{x\to c} h(x) = L$ it is understood that the limit actually exists and is $L$. Of course, write things like $\lim_{x\to c} h(x) = \infty$ means that $h$ doesn't have a limit and that $h(x)$ is arbitrarily large. – azif00 Apr 13 '20 at 02:53
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Both $L=\infty$ and $L=-\infty$ work fine. – Paramanand Singh Apr 13 '20 at 03:02
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My reading is that this remains valid. That is, it is within the range of the usual statement of Lhopital's rule. Always has been. Johann Bernoulli discovered it a long time ago, and it has endured. Thus, there will be no counter-example. For a proof, you can see the Wikipedia page, for instance. – Apr 13 '20 at 02:59