Showing a sequence is convergent using the $\epsilon$-$N$ definition I need to prove that the following sequence converges:
$\lim_{n\rightarrow \infty} \frac{2n^2+3n+1}{n^2+n+1}=2$
So for the proof/solution I have the following:
Let $\epsilon >0$. Then let $N=\frac{1}{\epsilon}$.  Then for all $n\geq N$,
$|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
Thus the sequence converges to 2.
Is this the correct way of going about this?  Thanks in advance.

There is a small error with symbols, it should be like this:
Choose $\epsilon\gt 0$
Let $N=\lceil\frac{1}{\epsilon}\rceil$.  Then for all $n\gt N$,
$|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$
