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,

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

${\left(1+x^2\right)}^n>1+n{x}^2$

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

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

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

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

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

$\Rightarrow L<{\tan }^{-1}\infty =\frac{\pi }{2}$

$\therefore \frac{1}{2}<L<2 \left[\because \frac{\pi }{2}<2\right]$