# How to FIND the limit of a sequence with epsilon definition?

So I have this sequence in hand: $x_n=\frac{\sqrt{n}}{n+1}$, and I can intuitively see that its limit as $n\to\infty$ is 0, and I can verify it with the $\varepsilon$ definition of the limit. But I am asked to FIND this limit with only the epsilon definition of the limit, how can I do that? (I can only find the limit by manipulating the statement into $\sqrt{\lim_{n\to\infty}{\frac{1}{n+2+\frac{1}{n}}}}$)

• The $\epsilon$-$N$ definition of limit is not a tool for finding the limit. – André Nicolas Nov 9 '14 at 2:49

Note that for any $\epsilon > 0$,
$$|x_n-0|=\left| \frac{\sqrt{n}}{n+1} - 0\right| < \frac{1}{\sqrt{n}} < \epsilon,$$
if $n > 1/ \epsilon^2.$
Hence $x_n \rightarrow 0$ and a limit when it exists is unique.