How to calculate $\lim_{n \to \infty} \int^{2007}_{0}e^{\frac{x^{2008}}{n}}dx$? How to calculate $$\lim_{n \to \infty} \int^{2007}_{0}e^{\frac{x^{2008}}{n}}dx?$$ Can I just write $e^{\frac{x^{2008}}{n}} \rightarrow e^0$ when $n \to \infty$?
 A: Note that $$\int_{0}^{2007}e^{x^{2008}/n}dx\leq\int_{0}^{2007}e^{x^{2008}}dx\leq2007e^{2007^{2008}}
 $$ so by the dominated convergence theorem $$\lim_{n\rightarrow\infty}\int_{0}^{2007}e^{x^{2008}/n}dx=\int_{0}^{2007}\lim_{n\rightarrow\infty}e^{x^{2008}/n}dx=2007.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&
2007 <
\int_{0}^{2007}\exp\pars{x^{2008} \over n}\,\dd x < 2007\exp\pars{2007^{2008} \over n}
\\[3mm]
&\imp\quad\color{#f00}{%
\lim_{n \to \infty}\int_{0}^{2007}\exp\pars{x^{2008} \over n}\,\dd x} =
\color{#f00}{2007}
\end{align}
A: We note that  $$e^{x^{2007}/n}= 1+ \frac{x^{2007}}{n}+ \frac{x^{4014}}{2n^2}+\dotsb $$ Since this series converges uniformly we can integrate term by term and we get that
$$\int_{0}^{2007} e^{x^{2007}/n}\, dx= \int_{0}^{2007}1 \, dx+ \int_{0}^{2007}\frac{x^{2007}}{n}\, dx+ \int_{0}^{2007}\frac{x^{4014}}{2n^2}\, dx+\dotsb $$  Letting $n$ go to infinity we see that all integrals except the first vanish in the limit. The result is  $2007$.
