Limit exercise from Rudin: $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$

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.

$\sqrt{n^2+n} - n= \frac{n}{\sqrt{n^2+n} + n} = \frac{1}{\sqrt{1+\frac 1n} + 1}$

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: $$ \sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}} $$