Calculate $\lim_{n\to{+}\infty}{(\sqrt{n^{2}+n}-n)}$ 
Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ 

Could someone help me through this problem? 
Calculate $\displaystyle\lim_{n \to{+}\infty}{(\sqrt{n^{2}+n}-n)}$
 A: We have:
$$\sqrt{n^{2}+n}-n=\frac{(\sqrt{n^{2}+n}-n)(\sqrt{n^{2}+n}+n)}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}$$
Therefore: 
$$\sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$
And since: $\lim\limits_{n\to +\infty}\frac{1}{n}=0$
It follows that:
$$\boxed{\,\,\lim\limits_{n\to +\infty}(\sqrt{n^{2}+n}-n)=\dfrac{1}{2}\,\,}$$
A: Guide: Rationalize,
$$\left(\sqrt{n^2+n}-\sqrt{n^2}\right)\cdot \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}}=\frac{n}{\sqrt{n^2+n}+\sqrt{n^2}}$$
Now divide numerator and denominator by $n$. Remember $\frac{1}{n}\sqrt{\square}=\sqrt{\frac{1}{n^2}\square}$.
A: Here's an answer that is probably not within the intended scope but it's nice anyway...
Let $x=1/n$. Then 
$$
\lim_{n\to{+}\infty}{\sqrt{n^{2}+n}-n} = \lim_{x\to0}{\sqrt{\frac1{x^2}+\frac1x}-\frac1x} = \lim_{x\to0}{\sqrt{\frac{1+x}{x^2}}-\frac1x} = \lim_{x\to0}{\frac{\sqrt{1+x}}{x}-\frac1x}= \lim_{x\to0}{\frac{\sqrt{1+x}-1}{x-0}} = f'(0) = \frac12
$$
for $f(x)=\sqrt{1+x}$.
(There's a small technicality that actually $x\to0^+$ but let's overlook that.)
