Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$

Question

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:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$

Answer 1

$\begin{align} \sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\ & = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\ & = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\ & = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\ \end{align}$

Answer 2

$$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$