Question: This is Chapter 3, Exercise 2 of Rudin's Principles.

Calculate $\lim\limits\_{n \to \infty} \sqrt{n^2+n} -n$.

Hints will be appreciated.

Answer: 
Hint:
$$\frac{\sqrt{n^2+n}-n}{1} = \frac{\sqrt{n^2+n}-\sqrt{n^2}}{1}\times \frac{\sqrt{n^2+n}+\sqrt{n^2}}{\sqrt{n^2+n}+\sqrt{n^2}} = \cdots$$
I will expand more if needed.
When you have a sequence that involves difference between two roots, it's often a good idea to try using the identity $(a-b)(a+b) = a^2-b^2$ to git rid of the root difference:
$$
\sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}
$$