TITLE: Calculate $\lim_{n\to{+}\infty}{(\sqrt{n^{2}+n}-n)}$
QUESTION [6 upvotes]: 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 $\displaystyle\lim_{n \to{+}\infty}{(\sqrt{n^{2}+n}-n)}$

REPLY [6 votes]: 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.)