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.
 A: Let 
\begin{align*}
I_n&=\int_0^\infty\frac{dx}{(1+x^2)^n}=\left.\frac{x}{(1+x^2)^n}\right|_{x=0}^\infty+\int_0^\infty\frac{2nx^2\,dx}{(1+x^2)^{n+1}}\\
&=2n\int_0^\infty\frac{dx}{(1+x^2)^n}-2n\int_0^\infty\frac{dx}{(1+x^2)^{n+1}}=2nI_n-2nI_{n+1}
\end{align*}
So we get
\begin{align*}I_{n+1}&=\frac{2n-1}{2n}I_n=\frac{2n-1}{2n}\frac{2n-3}{2n-2}I_{n-1}=\cdots\\
&=\frac{(2n-1)!!}{(2n)!!}I_1=4^{-n}\binom{2n}{n}\frac \pi 2
\end{align*}
By Stirling's formula,
$$\int_0^1\frac{dx}{(1+x^2)^{n+1}}=I_{n+1}+O(\tfrac 1 n)\sim \frac{1}{\sqrt{\pi n}}\cdot\frac\pi 2=\frac{\sqrt\pi}{2\sqrt n}$$
Thus the original limit equals
$$\sqrt n\int_0^1\frac{dx}{(1+x^2)^n}\sim\frac{\sqrt{\pi n}}{2\sqrt{n-1}}\to\frac{\sqrt \pi}{2}=0.8862\ldots$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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This can be evaluated by means of
Laplace Method:
\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{1}{\dd x \over \pars{1 + x^{2}}^{n}}}}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{1}
\exp\pars{-n\ln\pars{1 + x^{2}}}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{n^{1/2}\int_{0}^{\infty}
\exp\pars{-nx^{2}}\,\dd x}
\\[5mm] = &\
\int_{0}^{\infty}\exp\pars{-x^{2}}\,\dd x =
\bbx{\root{\pi} \over 2} \approx 0.8862 \\ &
\end{align}
A: 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
$$\lim_{n\to\infty}I_n = \frac{1}{2}\int_0^{\infty}\frac{du}{\sqrt{u}e^u}=\frac{1}{2}\int_0^{\infty}u^{-\frac{1}{2}}e^{-u}\;du$$ $$= \frac{1}{2}\Gamma\left(\frac{1}{2}\right) = \boxed{\frac{\sqrt{\pi}}{2}}$$
A: Nope, the limit cannot be zero. In a right neighbourhood of the origin $\frac{1}{1+x^2}\approx e^{-x^2}$, and for large values of $n$ we have that $\int_{0}^{1}e^{-nx^2}\,dx$ is horribly close to $\int_{0}^{+\infty}e^{-nx^2}\,dx$, which scales like $\frac{K}{\sqrt{n}}$ for a positive constant $K$. This actually is the main idea of the Laplace/Hayman methods. In our case
$$ \int_{0}^{1}\frac{dx}{(1+x^2)^n}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/4}\cos^{2n-2}(\theta)\,d\theta $$
is at most $\frac{1}{2^{n-1}}\cdot\frac{\pi}{4}$ apart from
$$ \int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{\pi}{2\cdot 4^{n-1}}\binom{2n-2}{n-1}=\frac{\pi n}{(2n-1)4^n}\binom{2n}{n}, $$
and since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$ (by Wallis product or similar elementary manipulations) we have
$$ \lim_{n\to +\infty}\sqrt{n}\int_{0}^{1}\frac{dx}{(1+x^2)^n}=\color{red}{\frac{\sqrt{\pi}}{2}}.$$
A: 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 $t>0$; 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 $$\lim_{n \to \infty} \int_0^{+\infty} \frac{1}{\left( 1 + \frac{t^2}{n} \right)^n} \chi_{[0,\sqrt{n}]}(t) \, dt = \int_0^{+\infty} e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}. $$
A: $$I(n)=\sqrt{n}\int_0^1\frac{dx}{(1+x^2)^n}$$
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
