The two sequences are $a_n = \frac{n}{n^2 + 1}$ and $s_n = \frac{1}{n} \sin(n)$.
I sort of know what do here
Obviously $\lim_{n\to\infty} a_n=0$, and you can do some sidework and say
$$\left|\frac{n}{n^2 + 1} - 0 \right| = \frac{|n|}{|n^2 + 1|} < \epsilon$$
Dont really know where to go from here.
I am wondering if n has to be a natural number.
We want to find $$\lim_{n\to\infty}\frac{n}{n^2+1}.$$ The limit is obviously $0$, and it seems to me that saying any more is a waste of time.
However, if we want to practice $\epsilon$-$N$ language, which can be useful elsewhere, we want to show that for ant $\epsilon\gt 0$ there is an $N$ such that if $n\get N$ then $$\left|\frac{n}{n^2+1}-0\right|\lt \epsilon.$$ Note that $0\lt \frac{n}{n^2+1}\lt \frac{1}{n}$. It follows that if $n$ is any integer greater than $\frac{1}{\epsilon}$, then $\frac{n}{n^2+1}\lt \epsilon$.
So we can take $N=\left\lceil \frac{1}{\epsilon}\right\rceil$, where $\lceil x\rceil$ is the ceiling function.
The second problem is essentially the same. The limit is clearly $0$, since $\sin(n)$ stays between $-1$ and $1$, and the $n$ at the bottom crushes it. Formally, we want to show that for every $\epsilon\gt 0$ there is an $N$ such that if $n\gt N$ then $$\left|\frac{\sin(n)}{n}-0\right|\lt \epsilon.$$ We observe that $\left|\frac{\sin(n)}{n}\right|\le \frac{1}{n}$ and finish like before.
For the first sequence $a_n$, note that
$$a_n=\frac{n}{n^2+1}=\frac{n/n}{(n^2+1)/n}=\frac{1}{n+1/n}.$$
So as $n\to\infty$, since $\frac{1}{n}\to0$, then $n+\frac{1}{n}\to\infty$, and thus
$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} \frac{1}{n+1/n}=0.$$
For the second sequence $b_n$, note that $0\leq|\sin(n)|\leq1$. As $n\to\infty$, $n$ is eventually positive, so:
$$0\leq \left|\frac{1}{n}\sin(n)\right|\leq \frac{1}{n}$$
Using the squeeze theorem, because $\frac{1}{n}\to0$ then $\left|\frac{1}{n}\sin(n)\right|\to0$ also, therefore $\lim_{n\to\infty}\frac{1}{n}\sin(n)=0$.