Limit exercise from Rudin: $\lim\limits_{n \to \infty} \sqrt{n^2+n} -n$ 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.
 A: $\sqrt{n^2+n} - n= \frac{n}{\sqrt{n^2+n} + n} = \frac{1}{\sqrt{1+\frac 1n} + 1}$
A: 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:
$$
\sqrt{a} - \sqrt{b} = \left(\sqrt{a} - \sqrt{b}\right)\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{\left(\sqrt{a} - \sqrt{b}\right)\left(\sqrt{a} + \sqrt{b}\right)}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{\sqrt{a} + \sqrt{b}}
$$
A: 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.
A: To get an idea of what the answer might be, I tried inputting values: n=100, n=1000, n=1000000.
e.g. for n = 1000,
sqrt(1000000 + 1000) ~ 1000 + 1/2 because (1000 + 1/2)^2 = 1000000 + 2*1000*(1/2) + 1/4 ~ 1000000 + 1000.
If you can spot this, then the rest is easy.
A: You can also do this by inductively showing it is increasing and bounded, but their ideas are probably a better route.
