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.