Calculate $\lim _{n\to \infty }a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx$

Hint. One may use an integration by parts, for $n\ge1$, $$ \begin{align} \int_0^1 x^{2n}\sin \frac{\pi x}{2}\:dx&=\left[\frac{x^{2n+1}}{2n+1}\cdot \sin \frac{\pi x}{2}\right]_0^1-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx \\\\&=\frac{1}{2n+1}-\frac{\pi}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx. \end{align} $$ Then observing that, as $n \to \infty$, $$ \left|\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx\right|\le\int_0^1 \left|x^{2n+1}\right|\:dx=\frac{1}{2n+2} \to 0, $$ one obtains, as $n \to \infty$, $$ n\int_0^1 x^{2n}\sin \frac{\pi x}{2}\:dx=\frac{n}{2n+1}-\frac{\pi\cdot n}{2(2n+1)}\int_0^1 x^{2n+1}\cos \frac{\pi x}{2}\:dx \to \frac12. $$ By writing, as $n \to \infty$, $$ a_n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx=\color{blue}{\frac{a_n}n} \cdot n\int _0^1 x^{2n}\sin \frac{\pi x}{2}dx $$one deduces an answer to the initial question.

Hint: A problem that has been here many times: If $f$ is continuous on $[0,1],$ then

$$\lim_{n \to \infty}(n+1)\int_0^1 x^n f(x)\,dx = f(1).$$

Use this with $f(x) = \sin (\pi x /2)$ and your result for $a_n.$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mbox{With Laplace Method,}\ \int_{0}^{1}x^{2n}\sin\pars{{\pi \over 2}\, x}\,\dd x = \int_{0}^{1}\pars{1 - x}^{2n}\cos\pars{{\pi \over 2}\, x}\,\dd x\sim {1 \over 2n}\ \mbox{as}\ n \to \infty \end{align}

So, you are left with

\begin{align} \lim_{n \to \infty}\pars{a_{n}\,{1 \over 2n}} = \bbx{\ds{1 \over \pi}} \end{align}