# 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 $N_\epsilon$)

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 ?

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}
• I choose $n_0$ such that ${n_0} > \sqrt {\frac{3}{\varepsilon }}$. – user226528 Mar 26 '15 at 5:41
• It's simpler if you go one step further and do $\frac{3}{n^2+1} < \frac{3}{n^2}$. – marty cohen Mar 26 '15 at 6:27
• @JamesMatthew marty cohen suggested working with $\frac 3{n_0^2}<\epsilon$ instead. This is much better than my original suggestion of working with $\frac 3{n_0^2+1}<\epsilon$. In fact, $\frac 3{n_0^2}<\epsilon$ will give you $n_0 > \sqrt{\frac 3{\epsilon}}$, as you had chosen precisely in the first place. – Cookie Mar 26 '15 at 6:50
Hint: $$\left | \frac{2n^{2}+\cos(n)}{n^{2}+1}-2 \right |=\frac{\left | -2+\cos(n) \right |}{n^{2}+1}\leq\frac{3}{n^{2}+1}<\epsilon.$$