Prove the limit of a sequence using definition Find the limit of the following sequence and justify it using the definition:
$$x_n = \frac{1}{n}sin\frac{nπ}{4}+\frac{n}{2n^2+cosn} $$
The definition:
X is the limit of the sequence $x_n$ if the following condition holds:
For each real number  Ɛ > 0, there exists a natural number N such that, for every natural number n > N, we have $|x_n - x|$ <  Ɛ.
I guessed the limit should be 0 and tried to evaluate $|x_n - 0|$ in many ways but I can't get it into a form where it's easy to see what n must be. 

You are correct, the limit is zero. To prove it by definition, let $\varepsilon > 0$ and set $N > \frac{2}{\epsilon}$. If $n > N$, then  
\begin{align}|x_n - 0| &= \left|\frac{1}{n}\sin \frac{n \pi}{4} + \frac{n}{2n^2 + \cos n}\right|\\
& \le \frac{1}{n}\left|\sin \frac{n\pi}{4}\right| + \left|\frac{n}{2n^2 + \cos n}\right|\\
& \le \frac{1}{n} + \frac{n}{2n^2 - 1} \\
& \le \frac{1}{n} + \frac{n}{n^2} \quad (\text{since $n^2 - 1 \ge 0$})\\
& = \frac{2}{n} < \frac{2}{N}\\
& < \epsilon
\end{align}
