L'Hôpital's rule only applicable if right-hand limit exist?

Question

In a source I have been reading, this statement was made regarding L'Hôpital's rule:

Why is it the case that L'Hôpital's rule is applicable only if the right-hand limit exists? Why not the left-hand? Why not both? I have read other sources on L'Hôpital's rule that do not mention it and would like clarification.

Here is the text:

L'Hôpital's Rule: Suppose that $f$ and $g$ are differentiable functions, and $f(a)=g(a)=0$, and suppose that $g'(x)$ is nonzero in a neighborhood of $a$ (except maybe at $a$ itself). Then $$\lim\limits_{x \to a} \ \frac{f(x)}{g(x)}=\lim\limits_{x \to a} \ \frac{f'(x)}{g'(x)}$$ if the limit on the right-hand side exists.

Answer 1

There are cases where $\lim\frac{f(x)}{g(x)}$ exists but $\lim\frac{f'(x)}{g'(x)}$ does not exist. For example, with $a=0$:

$$ f(x) = x^2\sin(1/x) \qquad\qquad g(x) = x $$

Here $\lim\limits_{x\to 0}\frac{f(x)}{g(x)}=0$, but $\frac{f'(x)}{g'(x)}$ does not have a limit for $x\to 0$.

Therefore L'Hospital's rule can only go in one direction: If ${f'(x)}/{g'(x)}$ happens to have a limit (and $f(x), g(x)$ both tend to $0$ or $\infty$), then this is also the limit of ${f(x)}/{g(x)}$.

But if ${f'(x)}/{g'(x)}$ does not exist, then this is not enough to conclude anything about whether ${f(x)}/{g(x)}$ has a limit or not.

Answer 2

My pocket example of such a limit is $$ \lim_{x \to \infty} \frac{x + \sin x}{x + \cos x}.$$ This limit is very clearly $1$. But an application of l'Hopital's rule would lead to the consideration of $$ \lim_{x \to \infty} \frac{1 + \cos x}{1 - \sin x},$$ which doesn't exist! Thus existence of the left hand limit does not guarantee the existence of the right hand limit.