Convergence of $x_n = \frac{n+\tan^{-1} n}{n^2 + \ln n}$ We know $\max\{\tan^{-1}\}=\frac{\pi}{2}.$ Therefore, for sufficiently large $n$, we have that $n>\tan^{-1}n.$ The limit of this sequence is zero. Therefore $$\frac{n+\tan^{-1} n}{n^2 + \ln n}\leq\frac{2n}{n^2 + \ln n}\leq\frac{2}{n}$$
and we can choose $N\geq \frac{2}{\epsilon}$ for the proof. Correct? I think the initial assumption is justified, since we are looking at the behaviour when $n\to \infty$.
 A: The basic idea behind your proof is solid and would readily convince most people. However, if this is for a class or some assignment, there is some sloppiness that might cause you to lose points. Let me restate your argument in a more air-tight manner:
Let $\epsilon>0$ be given and choose $N=\max\left\{\left\lceil\frac{2}{\epsilon}\right\rceil,2 \right\}$. Then for $n\geq N$ we have
$$\left|\frac{n+\tan^{-1}(n)}{n^2+\ln(n)}-0\right|=\frac{n+\tan^{-1}(n)}{n^2+\ln(n)}<\frac{n+\frac{\pi}{2}}{n^2}<\frac{n+\frac{4}{2}}{n^2}\leq\frac{n+n}{n^2}=\frac{2}{n}\leq \frac{2}{N}\leq \epsilon$$
Note the three things I tightened up:
$1)$ It's $\sup\{\tan^{-1}(n)\}=\frac{\pi}{2}$ (not maximum)
$2)$ I covered the case where $\epsilon$ is large (as it is all $\epsilon>0$) which would imply that $N\geq 2$
$3)$ I made sure that the argument is of the form: For all $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_n-L|<\epsilon$
But again, your reasoning is good and the argument is convincing, and the details I mentioned are so minor that they wouldn't effect anything outside of a class or assignment setting.
