Suppose the limit $L=\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^1\frac{1}{{\left(1+x^2\right)}^n}dx$ exists and is larger than $\frac{1}{2}$, then

We have, 

$L=\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^1\frac{dx}{{\left(1+x^2\right)}^n}$

We know that, ${\left(1+x^2\right)}^n>1+n{x}^2$

$\Rightarrow \frac{1}{{\left(1+x^2\right)}^n}<\frac{1}{1+n{x}^2}$

$\Rightarrow {\int }_0^1\frac{dx}{{\left(1+x^2\right)}^n}<{\int }_0^1\frac{1}{1+n{x}^2}dx$

$\Rightarrow {\int }_0^1\frac{1}{{\left(1+x^2\right)}^n}<\frac{1}{\sqrt{n}}{\left[{\tan }^{-1}\sqrt{n}x\right]}_0^1$

$=\frac{1}{\sqrt{n}}{\tan }^{-1}\sqrt{n}$

$L=\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^1\frac{1}{{\left(1+x^2\right)}^n}dx<\underset{n\to \infty }{\lim }\sqrt{n}\frac{1}{\sqrt{n}}{\tan }^{-1}\sqrt{n}$

$\therefore L=\frac{\pi }{2}$