calculus - How to calculate $lim_{xtoinfty}frac{x}{x-sin x}$?

I tried to solve
$$
\lim_{x\to\infty}\frac{x}{x-\sin x}.
$$

After dividing by $x$ I got that it equals to:
$$

\lim_{x\to\infty}\frac{1}{1-\frac{\sin x}{x}}.
$$ Now, using L'hopital (0/0) I get that
$$
\lim_{x\to\infty}\frac{\sin x}{x} = \lim_{x\to\infty}\cos x
$$
and the lim at infinity for $\cos x$ is not defined. So basically I get that the overall limit of
$$
\lim_{x\to\infty}\frac{x}{x-\sin x}
$$ is $1$ or not defined?