Finding $\lim_{n\to \infty}\sqrt n \int_0^1 \frac{\,dx}{(1+x^2)^n}$ $$ 
\lim_{n\to\infty} n^{1/2}
\int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0
$$
Is my answer correct?
But I am not sure of method by which I have done.

If you are fine with using the Gamma function (specifically $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$) and the dominated convergence theorem (DCT), then there is another way of finding your limit:


*

*Having in mind to get an $e$-function in the denominator, using the substitution $x^2=\frac{u}{n}$ you get
$$I_n = \sqrt{n}\int_0^1 \frac{1}{(1+x^2)^n} dx = \frac 12 \int_0^{\color{blue}{n}}\frac{du}{\sqrt u\left(1+\frac un\right)^n}$$

*Now, in order to apply the DCT, we need to bound the integrand properly:


$$I_n = \frac{1}{2}\int_0^{\color{blue}{1}}\frac{du}{\sqrt u\left(1+\frac un\right)^n} + \frac{1}{2}\int_{\color{blue}{1}}^{\color{blue}{n}}\frac{du}{\sqrt u\left(1+\frac un\right)^n}$$
$$ <  \frac 12 \int_0^{\color{blue}{1}}\frac{du}{\sqrt u} + \frac{1}{2}\int_{\color{blue}{1}}^{\color{blue}{n}}\frac{du}{\sqrt u(1+u)}$$
$$< 1 + \int_{\color{blue}{1}}^{\color{blue}{\infty}}\frac{du}{u^{\frac{3}{2}}} = 2$$
So, we are allowed to apply the DCT and get
$$\lim_{n\to\infty}I_n = \frac{1}{2}\int_0^{\infty}\frac{du}{\sqrt{u}e^u}=\frac{1}{2}\int_0^{\infty}u^{-\frac{1}{2}}e^{-u}\;du$$ $$= \frac{1}{2}\Gamma\left(\frac{1}{2}\right) = \boxed{\frac{\sqrt{\pi}}{2}}$$
