Prove that $\lim_{n \to \infty} \int_0^2 e^{ x^2 / n}\,{\rm d}x$ exists and evaluate it. I need to show that this limit exists and then evaluate it. It is from a section on uniform convergence of sequences. I know that if $f_n \rightarrow f$ uniformly and each $f_n$ is integrable, then I can bring the limit inside of the integral. I'm not sure if this will be the right way to approach this, and if it is how to show that it is all of those things that would allow me to bring the limit in. Looking for some advice. Thanks
 A: Hint.
If $0\le x\le 2$ then
$$1\le e^{x^2/n}\le e^{4/n}.$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#0000ff}{\large\lim_{n \to \infty}\int_{0}^{2}\expo{x^{2}/n}\,{\rm d}x}
&=
\lim_{n \to \infty}\pars{\root{n}\int_{0}^{2/\!\root{n}}\expo{x^{2}}\,{\rm d}x}
=
\lim_{n \to \infty}{\ds{\int_{0}^{2n^{-1/2}}\expo{x^{2}}\,{\rm d}x} \over n^{-1/2}}
\\[3mm]&=
\lim_{n \to 0}{\ds{\int_{0}^{2n}\expo{x^{2}}\,{\rm d}x} \over n}
=
\lim_{n \to 0}{\expo{\pars{2n}^{2}}\pars{2} \over 1}
=
\color{#0000ff}{\large 2}
\end{align}
