Answer
(a) $0$
(b) Does not exist.
Work Step by Step
(a) Use $\lim_{n\to\infty}\frac{1}{n}=0$, we have $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n}{n^2+4}=\lim_{n\to\infty}\frac{1/n}{1+4/n^2}=\frac{0}{1+0}=0$
(b) As $sec(n\pi)=\frac{1}{cos(n\pi)}$ and $cos(n\pi)$ oscillates in a range of $[-1,1]$ when $n\to\infty$, the limit $\lim_{n\to\infty}sec(n\pi)$ does not exist.
