What is the limit of $\lim\limits_{x→∞}\frac{\sin x}{x}$ How do I evaluate this limit: $$\lim\limits_{x\to \infty}\frac{\sin x}{x}$$
Is it $0$? If yes, how so?
 A: we can see that for $x>0$ we have
$$-\frac{1}{x}\le\frac{\sin x}{x}\le+\frac{1}{x}$$
then by squeeze theorem you can conclude that $\lim\limits_{x\to+\infty}\frac{\sin x}{x}=0$ since $\lim\limits_{x\to+\infty}-\frac{1}{x}=\lim\limits_{x\to+\infty}+\frac{1}{x}=0$
A: Hint: Use the squeeze theorem with two other limits: $\lim_{x \to \infty} 1/x$ and similarly $\lim_{x \to \infty} -1/x$, since $|\sin x| \leq 1$.
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: To answer this you can keep in mind:
$x/\infty=0$. 
$\sin(x)$ is never undefined.
Therefor, we can logically deduce that the limit $=0$
