Solving limits of a sine wave $\lim_{x\to+\infty}\frac{\sin(x)}{x}$ So I got this assignment. And I was wondering how is it possible to get a limit from a constantly changing formula. 
$$\lim_{x\to+\infty}\frac{\sin(x)}{x}$$
Can I only look in the domain $]0,2\pi[$?
 A: No, you cannot only analyze the problem within $]0, 2\pi[$, although $\sin$ is periodic.
Yes, the limit exists and $=0$. We have $x > 0$ only if 
$$
\bigg| \frac{\sin x}{x} \bigg| \leq \frac{1}{x};
$$
given any $\varepsilon > 0$,
we have $1/x < \varepsilon$ if $x > 1/\varepsilon$;
hence we have proved this: for every $\varepsilon > 0$, we have $x > 1/\varepsilon$ only if 
$$
\bigg| \frac{\sin x}{x} \bigg| < \varepsilon;
$$
that is, 
$$
\lim_{x \to +\infty}\frac{\sin x}{x} = 0.
$$
A: A possible approach is to take help of the following:


*

*$$\lim_{x\to+\infty}\frac{1}{x}=0$$

*$$-1\le\sin x\le1$$
A: You can't look only in $]0,2\pi]$, as this function is not periodic. The shortest way is to observe that  $\biggl|\dfrac{\sin x}x\biggr|\le\biggl|\mkern2mu\dfrac1x\mkern2mu\biggr|$, which tends to $0$ as $x$ tends to $\infty$.
A: First note that
$$-1\leq\sin x\leq 1$$
$$-\frac1{|x|}\leq\frac{\sin x}{|x|}\leq \frac1{|x|}$$
Using the squeeze theorem, we have
$$-\lim\limits_{x \to \infty} \frac1{|x|}\leq\lim\limits_{x \to \infty} \frac{\sin x}{|x|}\leq\lim\limits_{x \to \infty} \frac1{|x|}$$
As $x\to\infty$, we have $|x|=x$. So we can drop the absolute value bars
$$-\lim\limits_{x \to \infty} \frac1{x}\leq\lim\limits_{x \to \infty} \frac{\sin x}{x}\leq\lim\limits_{x \to \infty} \frac1{x}$$
$$0\leq\lim\limits_{x \to \infty} \frac{\sin x}{x}\leq 0$$
Therefore
$$\lim\limits_{x \to \infty} \frac{\sin x}{x} =0$$
