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 know that $\sin(x)$ will always be between $[-1,1]$ for $x \in \mathbb{R}$.
The denominator tends to $+\infty$ as $x \to \infty$. So we can conclude, since any member of $\mathbb{R}$ in the interval $[-1,1]$ divided by $+\infty$ that the limit is:
$$\frac{x\in[-1,1]}{\infty} = \boxed0$$
