$\frac{\int fg dx}{\int g dx}=f(0)$ We already know that
$$\lim_{n \rightarrow +\infty} \int_{-1}^1 (1-x^2)^n dx = 0$$
If we have $f(x) \in C[-1,1]$ then prove
$$\lim_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } = f(0)$$
My thought is $\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } - f(0) = \lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 [f(x)-f(0)](1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } $ so assume $f(0) = 0$. And $\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^1 f(x)(1-x^2)^n dx}{\int_{-1}^1 (1-x^2)^n dx } \le \lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^{-\delta} M(1-x^2)^n dx + \int_{-\delta}^\delta \epsilon (1-x^2)^n dx + \int_{\delta}^1 M(1-x^2)^n dx}{\int_{-\delta}^{\delta} (1-x^2)^n dx } $ where $M$ is the upper bounder of $|f(x)|$ and $\epsilon$ is small enough since $f(0)$. It suffice to show that $\lim\limits_{n \rightarrow +\infty} \frac{\int_{-1}^{-\delta} (1-x^2)^n dx}{\int_{-\delta}^\delta (1-x^2)^n dx } =0$.

You could observe (by calculus) that $g_n(x) = (1-x^2)^n / \int_{-1}^1 (1-u^2)^ndu$ is the density function for a random variable $X_n$ with expectation $0$ and with variance tending to 0 as $n\to\infty$. (This last by checking $\int_{-1}^1x^2g_n(x)dx\to0$, which is a Beta function  calculation.) Hence $X_n\to0$ in probability, and hence in distribution.  Hence $Ef(X_n)\to f(0)$, by the ``portmanteau theorem.''
