<s>[INST] 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.
 [/INST] For $n>0$ we have 
(1). $ \sqrt {n^2+n}-n=\frac {n}{\sqrt {n^2+n}+n}.$
(2). $ n=\sqrt {n^2}<\sqrt {n^2+n}<\sqrt {n^2+n+\frac {1}{4}}=n+\frac {1}{2}.$ 
(3). Therefore $ \frac {1}{2}-\frac {1}{8n+2}=\frac {n}{(   n+\frac {1}{2})   +n}<\frac {n}{\sqrt {n^2+n} +n}<\frac {n}{n+n}=\frac {1}{2}.$
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