Limit of $\frac{f'(x)}{g'(x)}$ & $g'(x) \neq 0$ in Hypotheses of L'Hospital's rule.

So on the Wiki page about l'Hospital's rule it tells me the conditions for when this rule works. If the functions in the quotient are $f(x)$ and $g(x)$, then both have to approach zero as x approaches c, or ± $\infty$; and the limit of $\frac{f'(x)}{g'(x)}$ must exist; and $g'(x)≠0$.

I'm not sure what the second point means. Must exist? I would have thought that means that $g'(x)≠0$ but then the third point covers this. Does it mean that the limit can't be ±$\infty$? As with $\lim\limits_{x\to 0}\frac{x}{x^2}$?

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Think about $\lim_{x \rightarrow \infty} \sin x$. Does this limit exist? –  ShawnD Feb 17 '12 at 17:47
In some contexts, when a limit goes to $\infty$ one says that it does not exist. But usually it means that the function does not approach a number. For example if you consider the function that takes the vlaue -1 in the rationals and 1 in the irrationals has no limit whatever $x$ you take. –  Chu Feb 17 '12 at 17:51
The other condition that I forgot to mention was that both $f(x)$ and $g(x)$ must be differentiable on an interval enclosing c. If c is $\infty$, can it be enclosed? –  Korgan Rivera Feb 17 '12 at 18:03
@Korgan: If $c$ is $\infty$, that means that there exists an $M\gt 0$ such that $f$ and $g$ are differentiable on $(M,\infty)$. The "neighborhoods of $\infty$" are intervals of the form $(M,\infty)$; similar for $-\infty$. –  Arturo Magidin Feb 17 '12 at 18:52

When we write things like $$\lim_{x\to a}h(x) = \lim_{x\to a}H(x)$$ we usually mean "if either limit exists, then they both do and they are equal; if either limit does not exist, then neither limit exists; if either limit does not exist and equals $\pm\infty$, then so does the other."
In L'Hopital's Rule, we want to write $$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\text{provided some conditions are met.}$$ But the problem is that the equality does not follow when the second limit does not exist; that is, it's possible for $f(x)$ and $g(x)$ to both go to $0$ or to $\pm\infty$, for $\lim\frac{f(x)}{g(x)}$ to exist, and for $\lim\frac{f'(x)}{g'(x)}$ to not exist.
For example (taken from Counterexamples in Calculus, by Sergiy Klymchuk), take $f(x) = 6x+\sin x$, $g(x) = 2x+\sin x$, and consider the limit as $x\to\infty$. We have that both $f(x)$ and $g(x)$ approach $\infty$; and the limit of $\frac{f(x)}{g(x)}$ can be computed directly: \begin{align*} \lim_{x\to\infty}\frac{f(x)}{g(x)} &= \lim_{x\to \infty}\frac{6x+\sin x}{2x+\sin x}\\ &= \lim_{x\to\infty}\frac{\frac{1}{x}(6x + \sin x)}{\frac{1}{x}(2x+\sin x)}\\ &= \lim_{x\to\infty}\frac{6 + \frac{\sin x}{x}}{2 + \frac{\sin x}{x}} \\ &= \frac{6}{2} = 3. \end{align*} However, $f'(x) = 6+\cos x$, $g'(x) = 2+\cos x$, and so $$\lim_{x\to\infty}\frac{f'(x)}{g'(x)} = \lim_{x\to\infty}\frac{6+\cos x}{2+\cos x}$$ does not exist: if $x$ is an even multiple of $\pi$, $x=2n\pi$, $n$ a positive integer, then $f'(2n\pi) = 7$, $g'(2n\pi) = 3$, so $\frac{f'(2n\pi)}{g'(2n\pi)} = \frac{7}{3}$. If $x$ is an odd multiple of $\pi$, then we have $f'((2n+1)\pi) = 5$, $g'((2n+1)\pi) = 1$, so $\frac{f'((2n+1)\pi)}{g'((2n+1)\pi)} = 5$. Since we can find values of $x$ arbitrarily large where $\frac{f'(x)}{g'(x)}$ is equal to $\frac{7}{3}$, and points with $x$ arbitrarily large where $\frac{f'(x)}{g'(x)}$ is equal to $5$, the limit cannot exist, and therefore $$\lim_{x\to\infty}\frac{f(x)}{g(x)}\text{ is not equal to }\lim_{x\to\infty}\frac{f'(x)}{g'(x)},$$ even though all hypothesis except the existence of the latter limit are satisfied.