<s>[INST] Difficult limit: $ \lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $ I found this question online

Find the limit:$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx $$

I've been told that I need to use the gamma function by converting $n^{3/2}=n\sqrt{n}$ and doing $u$ sub $u=x\sqrt{n}\rightarrow du=\sqrt{n} dx$
$$\lim_{n \rightarrow \infty} n^{3/2} \int _0^1 \frac{x^2}{(x^2+1)^n}dx =
\lim_{n \rightarrow \infty}  \int _0^{\sqrt{n}} \frac{nx^2\sqrt{n}}{(x^2+1)^n}dx =
\lim_{n \rightarrow \infty}  \int _0^{\sqrt{n}} \frac{u^2}{(\frac{u^2}{n}+1)^n}dx
$$
$$=\lim_{n \rightarrow \infty}  \int _0^{\sqrt{n}} \frac{u^2}{(\frac{u^2}{n}+1)^n}dx
$$
How does one proceed from here to get the gamma function?
 [/INST] Using the inequalities $x^2-\frac12 x^4\le \log(1+x^2)\le x^2$, we find that
$$\int_0^1 x^2e^{-nx^2}e^{-nx^4/2}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \int_0^1 x^2e^{-nx^2}\,dx\tag1$$
Then, enforcing the substitution $x\mapsto x/\sqrt n$ in $(1)$ reveals
$$\frac1{n^{3/2}}\int_0^{\sqrt n} x^2e^{-x^2}e^{-x^4/2n}\,dx\le \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx\le \frac1{n^{3/2}}\int_0^{\sqrt n} x^2e^{-x^2}\,dx\tag2$$
Finally, multiplying $(2)$ by $n^{3/2}$, letting $n\to \infty$, and applying the squeeze theorem yields the coveted limit
$$\lim_{n\to\infty}n^{3/2} \int_0^1 \frac{x^2}{(x^2+1)^n}\,dx=\frac{\sqrt \pi}{4}$$
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