Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$ 
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sqrt{n^2+n} + (n+\frac{1}{2})}$
And I'm not sure how to go on from here. Help would be appreciated.

Hint:
\begin{align}
\sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} 
\end{align}
and
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n} -\frac{1}{2}\right| = \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n}\right| = \frac{n}{2(\sqrt{n^2+n}+n)^2} \leq \frac{n}{8n^2}
\end{align}
