Answer
Converges to $1$.
Work Step by Step
Given: $a_n=nsin(\frac {1} {n})$
It can be re-written as
$a_n=\frac{sin(\frac {1} {n})}{\frac {1}{n}}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{sin(\frac {1} {n})}{\frac {1}{n}}$
Consider $\frac {1}{n}=x$ and $x \to 0$ as $n \to \infty $
Thus,
$=\lim\limits_{x \to 0} \frac {sin(x)}{x}$
Note that the limit is of the form $\frac{0}{0}$
Therefore, we can apply L-Hospital's Rule.
$=\lim\limits_{x \to 0} \frac {cos(x)}{1}$
$=1$
Therefore, the given sequence converges to $1$.
