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

Let $x = \tan(t)$. We then get that
\begin{align}
I(n) & = \int_0^{\infty} \left(\dfrac{2x}{1+x^2} \right)^n dx = \int_0^{\pi/2} \sin^n(2t) \sec^2(t) dt = 2^n \int_0^{\pi/2} \sin^n(t) \cos^{n-2}(t) dt\\
& = 4\int_0^{\pi/2}\sin^2(t) \sin^{n-2}(2t)dt \tag{$\star$}
\end{align}
Replacing $t$ by $\pi/2-t$, we get
$$I(n) = 4 \int_0^{\pi/2} \cos^2(t) \sin^{n-2}(2t) dt \tag{$\perp$}$$
Adding $(\star)$ and $(\perp)$, we get that
$$2I(n) = 4 \int_0^{\pi/2} \sin^{n-2}(2t)dt \implies I(n) = 2 \int_0^{\pi/2} \sin^{n-2}(2t)dt$$
I trust you can take it from here, using this post, which evaluates $\displaystyle \int_0^{\pi} \sin^{k}(t) dt$ and using Stirling (or) Wallis formula.
