## Calculus: Early Transcendentals 8th Edition

Converges to $1$.
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 $1$.