Prove $\lim_{n\to\infty}\frac{\int_{-1}^{1}{f(x)(1-x^2)^n}dx}{\int_{-1}^{1}{(1-x^2)^n}dx}=f(0)$ Prove that:Let $f\in C[-1,1]$
$$\lim_{n\to\infty}\frac{\int_{-1}^{1}{f(x)(1-x^2)^n}dx}{\int_{-1}^{1}{(1-x^2)^n}dx}=f(0)$$
My attempt:
$$\begin{eqnarray}\lim_{n\to\infty}\frac{\int_{-1}^{1}{f(x)(1-x^2)^n}dx}{\int_{-1}^{1}{(1-x^2)^n}dx}=&\lim_{n\to\infty}\frac{\int_{0}^{1}{f(x)+f(-x)(1-x^2)^n}dx}{2\int_{0}^{1}{(1-x^2)^n}dx}\\
&=\lim_{n\to\infty}\frac{(f(\xi)+f(-\xi))(\int_{0}^{1}{1-x^2)^n}dx}{2\int_{0}^{1}{(1-x^2)^n}dx}
\end{eqnarray}$$
But I can't complete this.
Any help would be greatly appreciated :-)
 A: Here is a standard argument dealing with this type of problem, often known as approximation-to-the-identity. Here is a slightly general claim, which does not harm the essence of the argument.

Proposition. Let $K_n \in C([-1,1])$ and assume that $(K_n)$ satisfy the following assumtions:

*

*Nonnegativity. $K_n(x) \geq 0$ for all $x \in [-1, 1]$ and $n \geq 1$.

*Normalization. $\int_{-1}^{1} K_n(x) \, \mathrm{d}x = 1$ for all $n \geq 1$.

*Concentration. For any $\delta > 0$, $\lim_{n\to\infty} \int_{|x|>\delta} K_n(x) \, \mathrm{d}x = 0$.

Then for any $f \in C([-1, 1])$, we have
$$ \lim_{n\to\infty} \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x = f(0). $$

In other words, $K_n(x)$ approximates the Dirac delta $\delta(x)$, which is "the identity" of the convolution (hence the name approximation-to-the-identity). Before the proof, let us check that this can be applicable to OP's problem. Indeed, set
$$K_n(x) = \frac{(1-x^2)^n}{\int_{-1}^{1} (1-t^2)^n \, \mathrm{d}t}. $$
Then the conditions 1 and 2 are obvious. For the concentration, notice that $(1 - t^2)^n \geq (1 - |t|)^n$, and so,
$$ \int_{|x|>\delta} K_n(x) \, \mathrm{d}t
\leq \frac{(1 - \delta^2)^n}{\int_{0}^{1}(1 - t)^n \, \mathrm{d}t} = (n+1)(1 - \delta^2)^n \xrightarrow[n\to\infty]{} 0. $$
Therefore $\int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x \to f(0)$ by the conclusion of the proposition.

Proof of Proposition. For any $\epsilon > 0$, find $\delta > 0$ so that $|x| < \delta$ implies $|f(x) - f(0)| < \epsilon$. Also, choose a bound $M > 0$ of $f$. Then
\begin{align*}
\left| \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x - f(0) \right|
&\leq \int_{-1}^{1} |f(x) - f(0)|K_n(x) \, \mathrm{d}x \\
&= \int_{|x|<\delta} |f(x) - f(0)|K_n(x) \, \mathrm{d}x + \int_{|x|\geq\delta} |f(x) - f(0)|K_n(x) \, \mathrm{d}x \\
&\leq \int_{-1}^{1} \epsilon K_n(x) \, \mathrm{d}x + \int_{|x|\geq\delta} 2M K_n(x) \, \mathrm{d}x \\
&= \epsilon + 2M \int_{|x|\geq\delta} K_n(x) \, \mathrm{d}x.
\end{align*}
So we have
$$ \limsup_{n\to\infty} \left| \int_{-1}^{1} f(x)K_n(x) \, \mathrm{d}x - f(0) \right| \leq \epsilon. $$
But since the left-hand side a fixed number independent of $\epsilon$ and the right-hand side can be made arbitrarily small, letting $\epsilon \downarrow 0$ proves the desired claim. $\square$
A: Let $x=y/\sqrt n.$ The expression becomes
$$\frac{\int_{-\sqrt n}^{\sqrt n}f(y/\sqrt n)(1-y^2/n)^n\,dy}{\int_{-\sqrt n}^{\sqrt n}(1-y^2/n)^n\,dy}.$$
Now $(1-y^2/\sqrt n)^n\le e^{-y^2}$ on $[-\sqrt n,\sqrt n].$ And $f$ is bounded. Also note that for fixed $y,$ $f(y/\sqrt n) \to f(0)$ by the continuity of $f$ at $0.$ It follows from the DCT that the limit is
$$\frac{\int_{-\infty}^{\infty} f(0)e^{-y^2}\, dy}{\int_{-\infty}^{\infty} e^{-y^2}\, dy} = f(0).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\lim_{n \to \infty}{\int_{-1}^{1}\mrm{f}\pars{x}
\pars{1 - x^{2}}^{n}\,\dd x \over
\int_{-1}^{1}\pars{1 - x^{2}}^{n}\,\dd x}} =
\lim_{n \to \infty}{\int_{0}^{1}
\bracks{\mrm{f}\pars{x} + \mrm{f}\pars{-x}}
\pars{1 - x^{2}}^{n}\,\dd x \over
2\int_{0}^{1}\pars{1 - x^{2}}^{n}\,\dd x}
\\[5mm] = &\
{1 \over 2}\,\lim_{n \to \infty}{\int_{0}^{1}
\bracks{\mrm{f}\pars{x} + \mrm{f}\pars{-x}}
\expo{n\ln\pars{1 - x^{2}}}\,\dd x \over
\int_{0}^{1}\expo{n\ln\pars{1 - x^{2}}}\,\dd x}
\\[5mm] = &\
{1 \over 2}\,\lim_{n \to \infty}{\int_{0}^{\infty}
\bracks{\mrm{f}\pars{0} + \mrm{f}\pars{-0}}
\expo{-nx^{2}}\,\dd x \over
\int_{0}^{\infty}\expo{-nx^{2}}\,\dd x}
\end{align}

The last step involves the
Laplace's Method which requires some $\ds{\mrm{f}\pars{x}}$ 'nice conditions' as you can see in the above link: I already assumed that's the case.

$$
\mbox{Then},\quad
\bbox[10px,#ffd]{\lim_{n \to \infty}{\int_{-1}^{1}\mrm{f}\pars{x}
\pars{1 - x^{2}}^{n}\,\dd x \over
\int_{-1}^{1}\pars{1 - x^{2}}^{n}\,\dd x}} = \bbx{\mrm{f}\pars{0}}
$$
