Calculate $\lim_{n\rightarrow\infty} \int_0^1 nx^2(1-x^2)^n \, dx$

Alternative approach: by setting $x=\sin\theta$ we have

$$ I(n) = \int_{0}^{1}nx^2(1-x^2)^n\,dx = n\int_{0}^{\pi/2}\sin^2(\theta)\cos^{2n+1}(\theta)\,d\theta \tag{1}$$ and since over the interval $\left[0,\frac{\pi}{2}\right]$ both $\sin(\theta)$ and $\cos(\theta)$ are non-negative, but $\sin(\theta)\leq\theta$ and $\cos(\theta)\leq e^{-\theta^2/2}$, it follows that: $$ 0\leq I(n) \leq \int_{0}^{\pi/2} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta \leq \int_{0}^{+\infty} n\theta^2 e^{-\left(n+\frac{1}{2}\right)\theta^2}\,d\theta=\sqrt{\frac{\pi}{2}}\frac{n}{(2n+1)^{3/2}}\tag{2}$$ and the wanted limit is zero by squeezing.

Since $1-x^2<e^{-x^2}$ for each $0\leq x\leq 1$, the integrand is non-negative and bounded above by $nx^2e^{-nx^2}$. Since $ye^{-y}$ is a bounded function on $[0,\infty)$, and $nx^2e^{-nx^2}\to 0$ for all $x\geq 0$, the result follows from the dominated convergence theorem.

Integrating by parts, $$ \int_0^1 nx^2(1-x^2)^n \, dx = \left[ -\frac{n}{2(n+1)}x(1-x^2)^{n+1} \right]_0^1 + \frac{n}{2(n+1)}\int_0^1 x(1-x^2)^{n+1} \, dx \\ = \frac{n}{2(n+1)}\int_0^1 (1-x^2)^{n+1} \, dx $$ The fraction converges to $1/2$, so now we need to look at the remaining integral. $(1-x^2)^{n+2} < (1-x^2)^{n+1} $, so the integrand decreases as $n \to \infty$. We can then use the monotone convergence theorem or similar to show the integral tends to zero, so the limit is zero.