How to calculate $\lim_{x\to\infty}\frac{x}{x-\sin x}$? 
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? 

With the sandwich theorem
$\dfrac{x}{x+1}\leq \dfrac{x}{x-\sin x}\leq \dfrac{x}{x-1} $
$\lim \limits_{x \to +\infty}\dfrac{x}{x+1}=1\quad \text{and}\quad \lim \limits_{x \to +\infty}\dfrac{x}{x-1}=1\implies \lim \limits_{x \to +\infty}\dfrac{x}{x-\sin x}=1 $
