On proving $\lim \frac{n^2}{n^2+n+1} = 1$ 
Prove that $\lim \frac{n^2}{n^2+n+1} = 1$

Let $\varepsilon > 0$ and let $N = \frac{1}{\varepsilon}.$ Then $n > N$ implies $n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon.$
But $\displaystyle \frac{1}{n} = \frac{n+1}{n(n+1)} = \frac{n+1}{n^2+n} > \frac{n+1}{n^2+n+1} = \bigg|\frac{n^2}{n^2+n+1}-1\bigg|.$
Therefore $\bigg|\frac{n^2}{n^2+n+1}-1\bigg| < \varepsilon$, hence $\displaystyle \lim \frac{n^2}{n^2+n+1} = 1.$
Could someone please verify whether the above is correct. 

if $\forall n.a_n \gt 0$
$$
\lim_{n\to\infty} a_n = 1 \Leftrightarrow \lim_{n\to\infty} \frac1{a_n} = 1
$$
set 
$$
a_n = \frac{n^2+n+1}{n^2} = 1+\frac1{n}+\frac1{n^2}
$$
so for $n \gt 1$ 
$$
0 \lt a_n-1 \lt \frac2n
$$
