TITLE: Finding limit for the function
QUESTION [7 upvotes]: I have problem with showing that the limit of the following function
$$\frac{
 \sqrt{\frac{3 \pi}{2n}} -
 \int_0^{\sqrt 6}(
 1-\frac{x^2}{6}
 +\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$
equal to $1$, with $n \to \infty$.

REPLY [6 votes]: Hint: First note that for $x\in[0,\sqrt{6}]$, $1-\frac{x^2}{6}+\frac{x^4}{120}$ monotonically decreases from $1$ to $\frac{3}{10}$ and that
$$
1-\frac{x^2}{6}+\frac{x^4}{120}\le1-\frac{x^2}{9}\le e^{-x^2/9}\tag{1}
$$
You might try the change of variables $x\mapsto x/\sqrt{n}$ so that
$$
\int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x
=\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\tag{2}
$$
and $(1)$ then says that for $x\in[0,\sqrt{6n}]$
$$
\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\le e^{-x^2/9}\tag{3}
$$
Thus, we have
$$
\begin{align}
\int_a^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x
&\le\int_a^\infty e^{-x^2/9}\;\mathrm{d}x\\
&\le\frac{9}{2a}e^{-a^2/9}\tag{4}
\end{align}
$$
Notice that the integrand on the right in $(2)$ tends to $e^{-x^2/6}$, so consider
$$
\begin{align}
\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n
&=e^{-x^2/6}\exp\left(n\log\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)+\frac{x^2}{6}\right)\\
&=e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}+O\left(\frac{x^8}{n^3}\right)\right)\tag{5}
\end{align}
$$
Applying $(5)$ to $(2)$ yields
$$
\begin{align}
&\left|\;\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x
-\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x\;\right|\\
&\le\frac{1}{\sqrt{n}}\int_0^{\log(n)} e^{-x^2/6}O\left(\frac{x^8}{n^3}\right)\;\mathrm{d}x\\
&+\frac{1}{\sqrt{n}}\int_{\log(n)}^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x
+\frac{1}{\sqrt{n}}\int_{\log(n)}^\infty e^{-x^2/6}\left|1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right|\;\mathrm{d}x\\
&=\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\tag{6}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x
&=\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\\
&=\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x+\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\\
&=\sqrt{\frac{3\pi}{2n}}\left(1-\frac{3}{20n}+\frac{11}{160n^2}+O\left(\frac{1}{n^3}\right)\right)\tag{7}
\end{align}
$$