Computing limits which involve square roots, such as $\sqrt{n^2+n}-n$ 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:
$\sqrt{n^2 + n} - n \lt \sqrt{n^2 +n + \frac{1}{4}} - n = n+\frac{1}{2} - n = 1/2$
 A: $$\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n =\lim_{n\rightarrow\infty}\sqrt{n^2 + n} - n\frac{\sqrt{n^2 + n}+n}{\sqrt{n^2 + n}+n}=...=\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n^2 + n}+n}=...=\frac{1}{2}$$  
A: $\begin{align}
\sqrt{n^2 + n} - n & = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} +n} \\
& = \frac{n^2+n-n^2}{\sqrt{n^2 + n} + n} \\
& = \frac{1}{\frac{\sqrt{n^2 + n} + n}{n}} \\
& = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} \rightarrow \frac{1}{2}\\
\end{align}$
A: By the bound stated in the problem, it suffices to show that the limit is $\ge \frac{1}{2}$. This proof is a lot worse than the others but maybe a good exercise in mean value theorem which is why I'm writing it up. Intuitively we want a way to estimate how far $\sqrt{n^2 + n}$ is from $n$. But the mean value theorem along with concavity gives exactly the approximation we need.
Let $f_n(x) = \sqrt{n^2 + x}$
$$f_n'(x) = \frac{1}{2}(n^2 + x)^{-\frac{1}{2}}$$ Now use the mean value theorem
$$\sqrt{n^2 + n} - n = f_n(n) - f_n(0) = f_n'(x^*)n$$  where $x^* \in (0, n)$ (and I have suppressed dependence of $x^*$ on $n$). Then by the fact that $f_n'$ is decreasing, we get
$$ f_n'(n) \lt f_n'(x*)$$
Multiplying by $n$ on both sides
$$ n f_n'(n) < n f_n'(x^*) = \sqrt{n^2 + n} - n$$
Substituting 
$$ \frac{1}{2}\frac{n}{(n^2 + n)^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$
$$ \frac{1}{2}\frac{1}{{(1 + 1/n)}^{\frac{1}{2}}} \lt \sqrt{n^2 + n} - n$$
And actually the upper bound of $1/2$ could have been found using the same method.
