A particularly tricky integral: $ \lim_{n \to \infty} n^{\frac{3}{2}} \int_0^1 \frac{x^2}{(1+x^2)^n} dx. $

By the change of variable $y = n^{1/2} x$ you get $$ n^{3/2} ∫_0^1\frac{x^2}{(1+x^2)^{n}}\,\mathrm{d}x = ∫_0^{n^{1/2}}\frac{y^2}{(1+n^{-1}y^2)^{n}}\,\mathrm{d}y \underset{n\to\infty}{⟶} ∫_0^\infty y^2e^{-y^2}\,\mathrm{d}y $$ by dominated convergence. Now, do the change of variable $x = y^2$ to get $$ ∫_0^\infty y^2e^{-y^2}\,\mathrm{d}y = \frac{1}{2}∫_0^\infty x^{1/2} e^{-x}\,\mathrm{d}x = \frac{\Gamma(3/2)}{2} = \frac{\Gamma(1/2)}{4} = \frac{\sqrt{\pi}}{4} $$

For large $n$ the integrand decreases rapidly away from $x=0$, so instead of integrating from $0$ to $1$ we may integrate from $0$ to $+\infty$. According to Maple $$ \eqalign{\int_0^\infty \frac{x^2}{(1+x^2)^n}\; dx &= \frac{\sqrt{\pi} \;\Gamma(n-3/2)}{4\; \Gamma(n)} \cr &\sim \frac{\sqrt{\pi}}{4} n^{-3/2} + \frac{15 \sqrt{\pi}}{32} n^{-5/2} + \ldots}$$