Prove $\lim_{x\to\infty}\sin x/x=0$ using definition of a limit of a function. Prove $\lim_{x\to\infty}\sin x/x=0$ using  definition of a limit of a function.
I know that the definition of a limit of a fuction $f(x)$ (when $x\to\infty$ and $f(x)$ has a finite limit) as: $\forall \epsilon >0$, $\exists M>0$ such that $x> M$, $|f(x)-L|<\epsilon .$ But, I dont know how to find $M$ (in terms of $\epsilon $) in order to prove $|\frac {\sin x}{ x}|<\epsilon$ ? Also we need to prove that using this definition, I stated above.
 A: First, note that what you must prove is that, for every $\varepsilon>0$ there is  a $M$ such that, for every $x>M$, $$\left|\frac{\sin x}{x}\right|< \varepsilon,$$
which is not quite what you wrote.
Note that $|\sin x |\le1$, therefore if $x>\frac{1}{\varepsilon}$, we will have the desired inequality.
A: You're a bit wrong with what you need to prove. In particular, since you want the limit to be $0$, you want $\left\lvert\frac{\sin x}{x}\right\rvert<\varepsilon$ and not $\left\lvert\frac{\sin x}{x}-1\right\rvert<\varepsilon$, which is what you would consider if you wanted the limit to be $1$.
Now let $\varepsilon>0$. Notice that, since $\lvert\sin x\rvert\leq 1$ for all $x\in\mathbb{R}$ we have that
$$\left\lvert\frac{\sin x}{x}\right\rvert\leq\frac{1}{\lvert x\rvert}$$
for all $x\in\mathbb{R}\setminus\{0\}$. Now suppose we have an $M>0$ and let $x>M$. Then the above inequality gives us that
$$\left\lvert\frac{\sin x}{x}\right\rvert<\frac{1}{M},$$
and so in particular if we choose $M=\frac{1}{\varepsilon}$, it follows from the above that
$$\left\lvert\frac{\sin x}{x}\right\rvert<\varepsilon$$
for all $x>M$, proving that
$$\lim_{x\to\infty}\frac{\sin x}{x}=0.$$
