Calculus - limit involving uniform convergence $$
\lim_{n\ \to\ \infty}\int_{0}^{\sqrt{\,n\,}\,}\left(1-{x^{2} \over n}\right)^{n}
\,{\rm d}x 
$$
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 !.
 A: 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: 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 $\frac{\sqrt{\pi}}{2}$.
A: $\newcommand{\+}{^{\dagger}}
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$\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}

