<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 each $\varepsilon>0$ you want to find $N$ such that, for $n>N$,
$$
\left|\sqrt{n^2+n}-n-\frac{1}{2}\right|<\varepsilon
$$
that is,
$$
-\varepsilon+\frac{1}{2}<\sqrt{n^2+n}-n<\varepsilon+\frac{1}{2}
$$
It is not restrictive to assume $0<\varepsilon<1/2$.
The inequality $\sqrt{n^2+n}-n>A$, for $A<1/2$, is the same as $\sqrt{n^2+n}>n+A$, and squaring gives
$$
n>2An+A^2
$$
that is $n>A^2/(1-2A)$. In the case of $A=(1-2\varepsilon)/2$ we have
$$
n>\frac{(1-2\varepsilon)^2}{8\varepsilon}
$$
The inequality $\sqrt{n^2+n}-n<B$, with $B>1/2$, is the same as $\sqrt{n^2+n}<n+B$ and, squaring,
$$
n<2Bn+B^2
$$
which is true for every integer $n$.
</s>