Prove using epsilon notation that $\lim_{n\to\infty} n\sin\frac {1}{n}=1$ Can someone help how to prove by epsilon notation $\lim_{n\to\infty} n\sin\frac {1}{n}=1$ ?
 A: You need to know that for any $x> 0$, 
$$x-\frac{x^3}6<\sin x<x\qquad\text{(this is easy to prove)}.$$
Setting $x=\dfrac1n$, we obtain:
$$1-\frac1{6n^2}<n\sin \frac1n<1,$$
whence
$$0<1-n\sin \frac1n<\frac1{6n^2}, \quad\text{so that}\quad\Bigl\lvert1-n\sin \frac1n\Bigr\rvert<\varepsilon\quad\text{if}\quad n>\frac1{\sqrt{6\varepsilon}}.$$
Proving the double inequality:
It is a consequence of Leibniz's theorem on alternating series, but it can be made an exercise for high school, based on the following corollary of the Mean value theorem:
Let $f, g$  be differentiable functions defined for $x\ge a$. If $f'(x)<g'(x) for $x>a$, then $f(x)-f(a)<g(x)-g(a)$ for $x>a$.

Apply the corollary to $\;\sin x< x\;$ for $x>0$. You obtain successively


*

*$0\le 1-\cos x<\dfrac{x^2}2$, whence $\;1-\dfrac{x^2}2<\cos x\le 1$,

*$x-\dfrac{x^3}6<\sin x \le x\;$ for $x>0$ (actually, we know the right-hand side inequality is strict).

A: You can rewrite your limit as 
$$
\lim_{x\to 0^+} \frac{\sin(x)}{x}.
$$
At this point, notice that $\sin(x)=x+O(x^2)$ (see, for example, here). This means that, if you fix $\varepsilon>0$, then you can find $\delta=\delta(\varepsilon)>0$ such that
$$
|x|<\delta \implies \sin(x) \in [x(1-\varepsilon),x(1+\varepsilon)].
$$
This implies that
$$
|x|<\delta \implies \frac{\sin(x)}{x} \in [1-\varepsilon,1+\varepsilon].
$$
Therefore your limit is $1$.
