# Computing limits which involve square roots

Is there a general strategy for this? For example I'm working on the limit
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n$$

I have a simple argument to show that this limit is less than or equal to 1/2, but I can't get much further because it's difficult for me to manipulate the square root symbol. Here is the argument:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$

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You can try a series expansion. –  Amzoti Feb 19 '13 at 0:06
Just write $\sqrt{n^2 + n} = n \sqrt{1 + \frac{1}{n}} \rightarrow n$ –  vonbrand Feb 19 '13 at 1:12

\begin{align} \sqrt{n^2 + n} - n & = \sqrt{n^2 + n} - n \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\ & = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\ & = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\ & = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\ \end{align}
@Mark: When in doubt, multiply by $1$, or add $0$. :) –  gnometorule Feb 19 '13 at 0:15
$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$