calculus - Why doesn't L'Hopital's rule work in this case?

I have a very simple question. Suppose I want to evaluate this limit:

$$\lim_{x\to \infty} \frac{x}{x-\sin x}$$

It is easy to evaluate this limit using the Squeeze theorem (the answer is $1$). But here both the numerator and the denominator are going to infinity as $x\to \infty$ so I tried using L'Hospital's rule:
$$\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}$$

However there's no finite $L$ such that $$\lim_{x\to \infty} \frac{1}{1-\cos x}=L$$
which is a contradiction. I don't understand why in this case L'Hopital's rule doesn't work. Both the numerator and the denominator are differentiable everywhere and both are tending to infinity - which is all we need to use this rule.

Answer

Another condition, very often forgotten, is that in some neighbourhood of $a$ (here $a=+\infty$), except perhaps at $a$, $g'(x)\neq 0$. This is not the case here: $1-\cos x$ is $0$ infinitely many times.

This is a good illustration why L'Hospital's rule is dangerous. One of the first things I learnt when I was a student is: ‘Avoid it. When it works, Taylor's polynomial at order 1 works as well.’

Here, the simplest way is via equivalents: $\,x-\sin x\sim_\infty x$ since $\sin x$ is bounded, hence
$$\frac x{x-\sin x}\sim_\infty \frac x{x}=1.$$