# How to find the limit of: [duplicate]

if the limit $$L=\lim_{n\to \infty}\sqrt n \int_0^1 \frac{1}{(1+x^2)^n}dx$$exists and is larger than $$\frac{1}{2}$$ then prove that $$\frac{1}{2} < L < 2$$ .

## marked as duplicate by StubbornAtom, Lee David Chung Lin, Lord Shark the Unknown, Jyrki Lahtonen, CesareoApr 14 at 8:58

Using $$(1+x^2)^n>1+nx^2$$ for $$x\geq 0$$
So $$\frac{1}{(1+x^2)^n}<\frac{1}{1+nx^2}$$
So $$\lim_{n\rightarrow \infty}\sqrt{n}\int^{1}_{0}<\lim_{n\rightarrow \infty}\sqrt{n}\int^{1}_{0}\frac{1}{1+nx^2}dx=\frac{\pi}{4}<2.$$
So we get $$\frac{1}{2}