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?
 A: You know that $-\dfrac{1}{x} \leq  \dfrac{\sin(x)}{x}  \leq \dfrac{1}{x}$
Now let $x \to \infty$ and apply the squeeze theorem.
A: If the limit of the numerator and denominator both existed, you could take the quotient of the limits. But they don't.
If the limit of the numerator and denominator were both $0$ or both infinite, then you could use L'Hospital's rule. But in fact the limit of the numerator doesn't exist at all.
Instead, this is easier to explain with the definition:
$$\left | \frac{\sin(x)}{x} - 0 \right | \leq \frac{1}{|x|}$$
so given $\varepsilon > 0$, if $x \geq \frac{1}{\varepsilon}$ then $\left | \frac{\sin(x)}{x} \right | \leq \varepsilon$. In other words you can make the ratio arbitrarily small in magnitude by taking $x$ sufficiently large.
If you prefer, you can invoke the squeeze theorem with the lower bound $-1/x$ and the upper bound $1/x$.
A: 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.
A: 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:
$$\frac{x\in[-1,1]}{\infty} = \boxed0$$
A: 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 $$\lim_{x \to \infty}\frac{\sin x}{x}=0$$
A: $\sin(x)$ is never undefined
and $\frac{\text{any number}}{\infty}=0$
Therefore you can deduce that it comes out to $0$
