Prove a sequence converges Show that the sequence $$p_n = n \sin\left(\frac 1 n\right)$$ converges to $1$ as $n$ goes to infinity. I am trying to use the epsilon delta definition of a limit to prove this. I know that $\sin(1/n)$ is bounded above by the function $1/n$. However, I cannot think of a lower bound for the function that will help to show convergence. Also, a step by step would be nice, to make sure I am writing my proof in the right form.
 A: You have $\lim_{x \to 0 } \frac{\sin x}{x} = 1$. This follows from l'Hôpital's rule, or a Taylor series expansion.
Since $\lim_{n \to \infty} \frac{1}{n} = 0$, we have $\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = 1$.
Addendum: To see the result using a series expansion:
The Taylor series for $\sin$ is $\sin x = x + \sum_{n=1}^\infty (-1)^{2n+1} \frac{x^{2n+1}}{(2n+1)!}$, hence for $x \neq 0$, we have
$\frac{\sin x}{x} = 1 + \sum_{n=1}^\infty (-1)^{2n+1} \frac{x^{2n}}{(2n+1)!}$.
For $|x|< 1$, we have $|1-\frac{\sin x}{x}| \le \sum_{n=1}^\infty (|x|^2)^n \le \frac{|x|^2}{1-|x|^2}$.
If you take $|x|<\frac{1}{2}$, you get $|1-\frac{\sin x}{x}| \le 2 |x|^2 \le |x|$, from which the desired result follows.
A: We will use the fact that if $0\lt x\le 1$ then
$$x-\frac{x^3}{6}\lt \sin x\lt x.$$
Putting $x=1/n$, we find that 
$$1-\frac{1}{6n^2} \lt n\sin(1/n)\lt 1.\tag{1}$$
Now we can use Squeezing. 
But if we want an $\epsilon$-$N$ proof of convergence, we continue as follows. From (1) we get
$$|n\sin(1/n)-1| \lt \frac{1}{6n^2}.$$
Now suppose that $\epsilon\gt 0$ us given. If $N$ the smallest  positive integer $\gt\dfrac{1}{\sqrt{6\epsilon}}$ and $n\ge N$ then 
$$|n\sin(1/n)-1|\lt \epsilon.$$
