$$ \lim_{n\ \to\ \infty}\int_{0}^{\sqrt{\,n\,}\,}\left(1-{x^{2} \over n}\right)^{n} \,{\rm d}x $$ I have a problem solving this limit...the only thing I was able to do is to prove that the function inside the integral uniformly converges to ${\rm e}^{-x^{2}}$.
Thanks !.
Define $f_n(x):=\chi_{(0,\sqrt n)}\left(1-\frac{x^2}n\right)^n$. Using the inequality $\log(1-t)\leqslant t$ for non-negative $t$, we get that $f_n(x)\leqslant e^{-x^2/2}$. Here, we have that $f_n\to e^{-x^2}$ uniformly on compact sets, so fix $\varepsilon>0$, and $A$ such that $\int_A^\infty e^{-x^2}\mathrm dx\lt\varepsilon$, then write for $n\gt A^2$, $$\left|\int_0^\infty f_n(x)\mathrm dx-\int_0^\infty e^{-x^2}\mathrm dx\right|\leqslant 2\varepsilon+\int_0^A|f_n(x)-e^{-x^2}|\mathrm dx.$$
A related problem. Using two changes of variables $y=\frac{x}{\sqrt{n}}$ and $z=y^2$ in a row puts the integral in the form of the beta function
$$ \int_0^ \sqrt{n} (1-(x^2/n))^n dx = \sqrt{n}\int_{0}^{1}( 1-y^2 )^n dy = \frac{\sqrt{n}}{2}\int_{0}^{1}z^{-1/2}( 1-z )^n dz $$
$$ = \frac{\sqrt{n}}{2}\frac{\Gamma(1/2)\Gamma(n+1)}{\Gamma(n+3/2)} .$$
Now, you can use the Stirling approximation
$$ n!=\Gamma(n+1)=n! \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}$$
to find the limit. The answer is $\frac{\sqrt{\pi}}{2}$.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\color{#c00000}{\tt Heuristically}$,
\begin{align} \int_{0}^{\root{n}}\pars{1 - {x^{2} \over n}}^{n}\,\dd x &=\int_{0}^{\root{n}}\exp\pars{n\ln\pars{1 - {x^{2} \over n}}}\,\dd x \\[3mm]&\sim\int_{0}^{\root{n}}\exp\pars{n\bracks{-\,{x^{2} \over n}}}\,\dd x \sim\int_{0}^{\infty}\expo{-x^{2}}\,\dd x = \color{#66f}{\large{\root{\pi} \over 2}} \end{align}
