Practice Preliminary exam - evaluate the limit This is from a practice prelim exam and I know I should be able to get this one.
$$
\lim_{n\to\infty} n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^n
$$
I have tried many different $u-$substitions but to no avail.  I have tried
$$
u = \log(1+x^2)
$$
$$
du = \frac{2x}{1+x^2}dx
$$
 but did not get anywhere

A related problem. Use the change of variables $ \frac{1}{1+x^2}=t $, then we have
$$ n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^ndx = \sqrt{n}2^{n-1} \int_0^1 t^{\frac{n}{2}-\frac{3}{2}}(1-t)^{\frac{n}{2}-\frac{1}{2}}dt$$
$$=\sqrt{n}2^{n-1}\beta\left(\frac{n}{2}-\frac{1}{2},\frac{n}{2}+\frac{1}{2}\right)=\sqrt{n} 2^{n-1}\frac{\Gamma(\frac{n}{2}-\frac{1}{2})\Gamma(\frac{n}{2}+\frac{1}{2})}{\Gamma(n)}=I(n), $$
where $\beta$ is the $\beta$ function. Taking the limit of $I(n)$, we have
$$ \lim_{n\to \infty} I(n)=\sqrt{2\pi}.  $$
You can use the Stirling' approximation $n!=\Gamma(n+1) \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n} $ of the gamma function to evaluate the limit 
