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 $\lim\limits_{x\to 0}\frac{x}{x^2}$?  
 A: 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.
