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L'Hopital's states the following:

Consider two functions $f(x),g(x)$, differentiable on an interval $U=(a,b)\subset\mathbb{R}$ such that $g'(x)\neq0$ $\forall x\in U$. If \begin{equation} \lim_{x\to c}f(x) = \lim_{x\to c}g(x) = 0 \end{equation} and \begin{equation} \lim_{x\to c}\frac{f'(x)}{g'(x)} = L, \end{equation} then \begin{equation} \lim_{x\to c}\frac{f(x)}{g(x)} = L. \end{equation}

My question is this - Are there alternative assumptions to $g'(x)\neq0$ $\forall x\in U$?

The wikipedia article states that there are (See the last paragraph of section 2. https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule). However, I do not have access to the references provided.

If this statement is correct, could someone formally state what these assumptions are and/or direct me to a source?

Thanks in advance!

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    The condition $g'(x) \neq 0$ is redundant because the existence of $\lim_{x\to c} \dfrac{f'(x)} {g'(x)} $ ensures that $g'(x) \neq 0$ as $x\to c$. – 2017-02-17
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    This may not be true. It can happen that the zeros of $f'(x)$ cancel with $g'(x)$, resulting in the limit of $f'(x)/g'(x)$ existing, whilst there may not be a neighbourhood of $c$ such that $g'(x) \neq 0$. – 2017-02-23
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    The usual definition of $\lim_{x\to a} f(x) $ assumes that $f$ is defined in some neighborhood of $a$ except possibly at $x=a$. So the case of vanishing denominator is not possible. – 2017-02-23
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    To clarify further, the usual definition of limits allows us to say that $\lim_{x\to 0}x/x=1$ (the usual case of cancelation) but the limit $\lim_{x\to 0}\dfrac{\sin(1/x)}{\sin(1/x)}$ does not exist because denominator vanishes as $x\to 0$. – 2017-02-23
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    Could you elaborate on this point? I dont understand how this excludes the possibility of infinitely many zeros. ie. $f(x) = x \sin (1/x) \to 0$ as $x\to 0$ but has infinitely many zeros. – 2017-02-23
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    The expression $\dfrac{\sin(1/x)}{\sin(1/x)}$ is not defined for infinitely many values near $x=0$ namely at points $x=1/n\pi$ because denominator vanishes at these points. The case of $x\sin(1/x)$ does not have same problem because there is no vanishing denominator here. – 2017-02-23
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    I think I understand your point. Just to clarify, consider the following: In the case that $f = g = x\sin (1/x)$, both $f, g$ tend towards $0$ as $x\to 0$. However, $\lim_{x\to 0}\frac{f'(x)}{g'(x)}$ does not exist as the denominator vanishes infinitely many times near $0$. As this limit does not exist, L'Hopital's rule does not apply. Is this correct? – 2017-02-23
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    Yes if the limit of $f'/g'$ does not exist (because of any reason whatsoever) then the rule does not apply. – 2017-02-23

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These are the two pages from Krantz:

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