Prove that $\lim\limits_{n\to\infty}\frac{n^2+n+1}{(n+1)^2}=1$ So, here's the sequence $\{x_n\}$ defined by the following formula:
$$x_n = \frac{n^2+n+1}{(n+1)^2}$$
I want to try and prove this from the definition. Let $\epsilon > 0$ be given. Then, we need an integer $N(\epsilon) > 0$ such that:
$$n > N \implies |x_n - 1| < \epsilon$$
$$|x_n-1| = |\frac{n^2+n+1}{(n+1)^2} - 1| = |\frac{n}{(n+1)^2}| < \frac{1}{n}$$
Then, define $N(\epsilon) = [\frac{1}{\epsilon}]+1$, where $[x]$ is the integer part of $x$. Since we have our required $N(\epsilon)$, this proves the desired result. 
Does the proof above work? If it doesn't, why and how can I fix it?

Yes your proof seems to be good. (the answer practically ends here, but here is another way to prove it)  
Proof. $$\frac {n^2+n+1} {(n+1)^2} = \frac {n^2+n+1} {n^2+2n+1}= 1 - \frac n {n^2+2n+1}$$.
We claim that $$\lim_{n\to+\infty}\frac n {n^2+2n+1} = 0$$, and it can be shown that for any $\epsilon \gt 0$, as long as $n \gt N = \frac 1 {\epsilon}$, $$\lvert \frac n {n^2+2n+1} \rvert \lt \epsilon$$.  It is also true that $$\lim _{n\to+\infty} 1 = 1$$
Therefore $$\lim_{n\to+\infty} \frac {n^2+n+1} {(n+1)^2} = 1$$.
