a) Proof: Given arbitrary $\displaystyle \epsilon > 0$, choose $\displaystyle N$ such that both these inequalities hold: $\displaystyle N > 100$ and $\displaystyle N > \frac{1}{\epsilon}$. Then we know $\displaystyle \forall n \geq N$, $\displaystyle x_n = 1/n < \epsilon$. Hence $\displaystyle x_n \rightarrow 0$.
b) Follows pretty easily from a. you should be able to handle it.