Can someone please check if the following proof of $\lim_{n\to\infty} \frac{n}{n^2 + 1} = 0$ is correct?
Let $\epsilon > 0$ and choose $N = \lceil1/\epsilon\rceil + 1$. Then for all $n > N$, we have
$$\left| \frac{n}{n^2 + 1} - 0\right|$$
$$= \frac{n}{n^2 + 1} \leq \frac{n}{n^2} = \frac{1}{n}.$$
But since $n > N = \lceil{1/\epsilon}\rceil + 1$, we have $n > 1/\epsilon$; or equivalently, $\epsilon > 1/n$. Therefore,
$$\frac{1}{n} < \epsilon,$$
as desired.
Your proof is correct except a typo in $$ \frac{n}{n^2 + 1} \leq \frac{n}{n} \leq \frac{1}{n}$$
Please fix it to $$\frac{n}{n^2 + 1} < \frac{n}{n^2} = \frac{1}{n}$$
and go from there.