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.
No. With the change of variable $ t = \sqrt{n} x $ you get $$ \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} \, dx = \int_0^\sqrt{n} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \, dt = \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt. $$Observe that $$ \lim_{n \to \infty} \left( 1 + \frac{t^2}{n} \right)^n \chi_{[0,\sqrt{n}]}(t) = e^{t^2}$$pointwise everywhere, say for $t&gt;0$; also the sequence $ n \mapsto (1 + t^2 /n)^n$ is increasing for all $t \in \mathbb{R}$ which implies that $$\frac{1}{1+t^2} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \ge \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) $$ for all $n \in \mathbb{N}$ and for all $t \in \mathbb{R}$. By the Dominated Convergence Theorem, since $1/(1+t^2) \in L^1([0,+\infty))$, we have that $$\lim_{n \to \infty} \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt = \int_0^{+\infty} e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}. $$
$$I(n)=\sqrt{n}\int_0^1\frac{dx}{(1+x^2)^n}$$
and since $$(1+x^2)^{-n}=\sum_{k=0}^\infty{{n+k-1}\choose k}(-1)^kx^{2k}$$
$$I(n)=\sqrt{n}\int_0^1\sum_{k=0}^\infty{{n+k-1}\choose k}(-1)^kx^{2k}dx=\sqrt{n}\sum_{k=0}^\infty{{n+k-1}\choose k}\frac{(-1)^k}{2k+1}$$
and we know that:
$${{n+k-1}\choose k}=\frac{(n+k-1)!}{k!(n-1)!}$$
and as $n\to\infty$ this will $\to1$
so we are left with:
$$I(n)\approx\sqrt{n}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$
now take the limit