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? 
 A: By the squeeze theorem, we have that $$|\sin(x)|\leq 1\implies \frac{-1}{x}\leq\frac{\sin(x)}{x}\leq\frac{1}{x}\implies \lim_{x\to\infty}\frac{\sin(x)}{x}=0$$
since $\lim_\limits{x\to\infty}\frac{1}{x}=0$. You may be thinking of $\lim_\limits{x\to 0}\frac{\sin(x)}{x}=1.$
A: L'Hopital is used in a wrong way. It is not the 0/0 type. 
Note that
$$
\lim_{x\to\infty}\frac{\sin x}{x}=0
$$
since $x\mapsto\sin x$ is bounded and 
$$
\lim_{x\to\infty}\frac{1}{x}=0.
$$
A: Note that
$$
\left\lvert\frac{\sin x}{x}\right\rvert\le\frac{1}{x}\to0 \quad {\text{as}\quad x \to \infty}.
$$
Hence
$$
\lim_{x\to\infty}\frac{\sin x}{x}=0
$$
and your limit is
$$
\frac{1}{1-0}=1
$$
A: 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 $
A: You have to do more work to justify applying L'Hopital's rule to pieces of expressions.  The short version is that since the little limit you evaluated does not exist, the manipulations to get there are invalid.  (Note that nearly every limit law says "this limit is equal to that expression containing potentially simpler limits as long as those simpler limits exist".  When first learning about limits, many students let that italicized phrase fall out of their memories, but it is a very important phrase.)
Also, $\frac{\sin x}{x}$ is not a $(0/0)$ indeterminate form as $x \rightarrow \infty$.  Neither $\sin x$ nor $x$ are going to zero.
Your limit can be attacked many ways.  Here's one \begin{align*}
\lim_{x \rightarrow \infty} \frac{x}{x- \sin x} 
    &= \lim_{x \rightarrow \infty} \frac{x - \sin x + \sin x}{x- \sin x} \\
    &= \lim_{x \rightarrow \infty} \left( \frac{x - \sin x}{x- \sin x} +\frac{\sin x}{x- \sin x}  \right)\\
    &= \lim_{x \rightarrow \infty} \left( 1 + \frac{\sin x}{x- \sin x}  \right) \\
    &= 1 + \lim_{x \rightarrow \infty} \frac{\sin x}{x- \sin x}  &&\text{(if the new limit exists) using sum and constant laws}  \\
    &= 1 + 0  &&\text{bounded numerator, unbounded denominator}  \\
    &= 1  \text{.}
\end{align*}
A: TLDR:
$$\frac 1L=\lim_{x\to\infty}\frac{x-\sin(x)}{x}=\lim_{x\to \infty}1-\frac{\sin(x)}{x}=1$$

Explanation:
Let $$f(x)=\frac{x-\sin(x)}{x}.$$
 Then 
$$\lim_{x\to \infty}f(x)=\lim_{x\to \infty} 1-\frac{\sin(x)}{x}=1.$$
This implies the existence of an $M>0$ such that $f(x)\in (\frac 12, \frac 32)$ for every $x>M$. Since $f(x)>0$ for $x>M$, the function $g(x)=\frac{1}{f(x)}$ is well defined for $x>M$ and has limit
$$\lim_{x\to \infty}\frac{x}{x-\sin(x)}=\lim_{x\to \infty} g(x)=\lim_{x\to \infty}\frac{1}{f(x)}=1$$
