Originally Posted by nubshat
Note that for all x greater then 1

$x-1 \le x+\sin(x) \le x+1 \iff \frac{1}{x+1} \le \frac{1}{x+\sin(x)} \le \frac{1}{x-1}$

and note that $-1 \le \sin(2x) \le 1$

Putting these two facts together gives

$\frac{-1}{x+1} \le \frac{\sin(2x)}{x+\sin(x)} \le \frac{1}{x-1}$

Now what happens as you take the limit?