Evaluating and proving $\lim\limits_{x\to\infty}\frac{\sin x}x$ I've just started learning about limits. Why can we say $$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 0 $$ even though $\lim_{x\rightarrow \infty} \sin x$ does not exist? 
It seems like the fact that sin is bounded could cause this, but I'd like to see it algebraically. 
$$ \lim_{x\rightarrow \infty} \frac{\sin x}{x} = 
\frac{\lim_{x\rightarrow \infty} \sin x} {\lim_{x\rightarrow \infty} x}  
= ? $$
L'Hopital's rule gives a fraction whose numerator doesn't converge. What is a simple way to proceed here?

We have $$\lim\frac{f}{g}=\frac{\lim f}{\lim g} $$
if all three limits exist and $\lim g$ is not zero. This doesn't mean that the existence of th elimit on the left would imply the existence of both limits on the right! 
Similarly, note for the application of l'Hopital that some conditions must be met: The limit $\lim\frac fg$  must be an "indeterminate form" and the limit of $\lim\frac{f'}{g'}$ must exist; neither is the case here.
