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
Probably not how you want this done, but a useful and elegant method nonetheless:
Note that $$f(x)=\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\to\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$.
$f^n \equiv \exp(n \log f)$ and $n\log(f(1+u))=-n\frac{1}{2}u^2+n\mathcal O(u^3)$. Then letting $v=\sqrt{n} u$, $$\int_0^\infty n^{1/2}\mathrm d x \exp(-\frac{1}{2}v^2+\mathcal O(n^{-1/2})) \approx \int_{-\infty}^\infty \mathrm d v\, e^{-v^2/2} = \sqrt{2\pi}$$
Verification: WolframAlpha.
Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals. (It's also how Stirling's approximation is most easily derived! Here one applies the method to the $\Gamma$ function.) It can be tightened up and made rigorous straightforwardly by estimating errors. You can also derive an asymptotic expansion by considering the correction terms.
One useful insight it offers is why the $\sqrt{n}$ is really there: it's because the peak contributing the dominant term has a width decaying as $1/\sqrt{n}$.
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.
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
$$\lim_{n\to\infty} n^{1/2}\int_0^\infty \left( \frac{2x}{1+x^2} \right)^n dx = {1\over\sqrt{n}}\lim_{n\to \infty} \int_0^{\pi/2} \left(2\tan(x)\over \sec^2(x)\right)^n{ \sec^2(x)\,dx}$$ Can you use this?
