<s>[INST] Proof : Limit of a sequence 
Prove from the definition of the limit of a sequence that $$\lim_{n\to\infty} \frac{2n^2+\cos(n)} {n^2+1} = 2 $$ 
(that is, for a given $\epsilon > 0$, find an explicit $N_\epsilon$)

Please explain.
I choose $n_0$ such that ${n_0} > \sqrt {\frac{3}{\varepsilon }} $. Then how can I say that for all $n\ge n_0$, we have $\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} - 2} \right| <\varepsilon$. ? How to proceed further ?
 [/INST] Let $\epsilon > 0$ be given. Choose an $n_0 \in \mathbb{N}$ such that $n_0 > \sqrt{\frac 1{\epsilon}}$. Then for all $n \in \mathbb{N}$ such that $n \ge n_0$, we have
\begin{align}
\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} -2 } \right|&=\left| {\frac{{2{n^2} + \cos n}}
{{{n^2} + 1}} - \frac{2(n^2+1)}{n^2+1}} \right|\\ &=\left|\frac{\cos n -2}{n^2+1} \right| \\
&=\left|\frac{\cos n}{n^2+1}-\frac 2{n^2+1} \right| \\
&\le\left|\frac{\cos n}{n^2+1} \right|+\left|\frac 2{n^2+1} \right| \\
&=\frac{|\cos n|}{n^2+1}+\frac 2{n^2+1} \\
&\le \frac 1{n^2+1}+\frac 2{n^2+1} \\
&=\frac 3{n^2+1} \\
&< \frac 3{n^2} \\
&\le \frac 3{n_0^2} \\
&<\epsilon
\end{align}
</s>