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?

From the boundedness of sinusoidal curves we know that
$$|\sin x \  | \le 1$$
hence
$$\bigg|\frac{\sin x}{x}\bigg| \leq \frac{1}{x}$$
Using our rules for absolute inequalities, we find that $$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$$
Now $$\lim_{x \to \infty} -\frac{1}{x} = 0 = \lim_{x \to \infty} \frac{1}{x} $$
If you are unsure as to why this is true, consider the following plot of $\displaystyle f(x) = \frac{1}{x}$. The same argument for $\displaystyle f(x)= - \frac{1}{x}$, since this is simply a reflection in the $x$-axis.
(Note: I plotted it only for positive $x$)

Thus, from the Squeeze Theorem, we thus have that $$\lim_{x \to \infty}\frac{\sin x}{x}=0$$
