Show, using the $N$, $\epsilon$ definition of convergence, that the sequence $\frac{2}{\sqrt{n+2}}$ converges to $0$. I have tried to do the following: let $\epsilon >0$, and we want to find a positive integer $N$ such that for all $n>N$, we have that $|2/(\sqrt{n+2})-0|$ and this should be less than $\epsilon$, but I'm not sure where to go from there.
 A: From the Archimedian property, you can always find $n_0\in\Bbb N$ such that $1/n_0<\epsilon/2$. Take $n_0>1$.  Then $\forall m\ge n_0^2-2,\sqrt{m+2}\ge n_0\implies \frac1{\sqrt{m+2}}\le\frac1{n_0}<\epsilon/2$.
A: Hint: write down the inequation, and solve it
Step by step:
$\left\vert \dfrac{2}{\sqrt{n+2}} \right\vert < \varepsilon$
$\dfrac{2}{\sqrt{n+2}} < \varepsilon$
$\sqrt{n+2} > \dfrac{2}{\varepsilon}$
$n > \left( \dfrac{2}{\varepsilon} \right)^2 - 2$
So take $N = \left\lceil \left( \dfrac{2}{\varepsilon} \right)^2 - 2 \right\rceil$, ensuring your above inequality for all $n \geq N$.
A: Usually in problems like these it helps to first 'work backwards' And only then write out the formal proof. For your problem, the 'backwards work' looks like
$$\epsilon>\left|\frac{2}{\sqrt{n+2}}-0\right|=\frac{2}{\sqrt{n+2}}$$
$$\sqrt{n+2}>\frac{2}{\epsilon}$$
$$n+2>\frac{4}{\epsilon^2}$$
$$n>\frac{4}{\epsilon^2}-2$$
Thus, choose $N=\max\left\{\frac{4}{\epsilon^2},1\right\}$. First, if $\epsilon\geq 2$, then $N=1$. This implies $n\geq N=1$ and therefore
$$n\geq 1$$
$$n+2\geq 3$$
$$\sqrt{n+2}\geq \sqrt{3}>1$$
$$\left|\frac{2}{\sqrt{n+2}}-0\right|=\frac{2}{\sqrt{n+2}}< 2\leq \epsilon$$
and we are done. Now, if $\epsilon<1$ then $N=\frac{4}{\epsilon^2}$. Thus $n\geq N=\frac{4}{\epsilon^2}$ implies
$$n\geq N=\frac{4}{\epsilon^2}>\frac{4}{\epsilon^2}-2$$
$$n+2>\frac{4}{\epsilon^2}$$
$$\epsilon^2>\frac{4}{n+2}$$
$$\epsilon>\frac{2}{\sqrt{n+2}}=\left|\frac{2}{\sqrt{n+2}}-0\right|$$
We conclude that
$$\lim_{n\to\infty}\frac{2}{\sqrt{n+2}}=0$$
