Proof of $\lim_{x \to \infty} \frac{\sin x}{x} =0$ Prove that $$\lim_{x \to \infty} \frac{\sin x}{x} =0$$ 

For a given $\epsilon \gt 0$ we have

\begin{align}\left |\frac{\sin x}{x} -0\right|&amp;\lt \epsilon\\
\implies \frac{|\sin x|}{|x|} &amp;\lt \epsilon\end{align}

From here how can we get an $M \gt 0$ such that $x \gt M$ $\implies$ $|f(x)-L|\lt \epsilon$
Well, the proof goes like this:

Let $\epsilon&gt;0$. We have to find a $M=M(\epsilon)&gt;0$, such that for every $x&gt;M$, we have that $\left|\frac{\sin x}{x}\right|&lt;\epsilon$.

Note that $|\sin x|\leq1$ for every $x\in\mathbb{R}$. Note also that, from Archimedes-Eudoxus Principle, we can find a $n_0=n_0(\epsilon)\in\mathbb{N}$ such that:
$$\frac{1}{n_0}&lt;\epsilon$$

Let $M=n_0$. Now, since $\frac{1}{x}$ is strictly decreasing and, hence $x&gt;M\Rightarrow\frac{1}{x}&lt;\frac{1}{M}=\frac{1}{n_0}&lt;\epsilon$, we have, for every $x&gt;M&gt;0$:
$$\left|\frac{\sin x}{x}\right|\leq\left|\frac{1}{x}\right|\overset{x&gt;0}{=}\frac{1}{x}&lt;\frac{1}{M}&lt;\epsilon$$
So, the proof is complete.
The function $\sin x$ always lies between $0$ and $1$. 
So it will have any of the value between $0-1$, when $x$ is tending to infinity.
So as $x$ tends to infinity $\frac{\sin x}{x}$ will be zero.