Showing convergence using $\epsilon$-$N$ definition? $$\lim_{n \to \infty} \frac{2}{\sqrt{n+3}}$$
Using the $\epsilon$-N definition:
Fix some $\epsilon>0.$
Let $N = \frac1\epsilon$.
Where I'm confused, since this series converges to 0, would you continue by stating that $|A - 0| < \frac1n$ and continue from here, or in an analysis class would this be wrong?
 A: You can also do it by contradiction. Suppose that there exists $\epsilon>0$ such that for every $N\in\mathbb N$ 
$|0-{2\over {\sqrt{n+3}}}|$=$2\over {\sqrt{n+3}}$$>\epsilon$
For that $\epsilon$ there exists $m\in\mathbb N$ such that ${{1\over {m+1}}<\epsilon\leq{1\over m}}$ so we must have ${2\over {\sqrt{n+3}}}>{1\over {m+1}}$ for some $m\in\mathbb N$ and every $n\in\mathbb N$. Going further we get $2(m+1)>\sqrt{n+3}$. Choosing $n\geq 4m^2+8m+1$ we obtain a contradiction so the limit exists and the limit is $0$.
A: Actually no. The $\epsilon-N $ definition would be used as follows
Fix $\epsilon >0$. Show that:
$$\exists N \quad \mbox{such that} \quad n\geq N \Rightarrow |A_n -0| < \epsilon$$
Where $A_n$ is the sequence provided. 
Here N can depend on $\epsilon$. Also to get some insight into what is actually happening, you are in effect saying that for large enough n, you are getting as close to the limit value as you want, even if you never reach it!
Hope this answers your query. 
Some would define it like this also:
Fix $m \geq 1$. Show that:
$$\exists N \quad \mbox{such that} \quad n\geq N \Rightarrow |A_n -0| < 1/m$$
