Evaluation of $\lim_{n\to\infty} \int_0^1 \frac{e^{\displaystyle x^{n}}}{1+x^2}\,\mathrm{d}x$ Evaluation of 
$$\lim_{n\to\infty} \int_0^1 \frac{e^{\displaystyle x^{n}}}{1+x^2}\,\mathrm{d}x$$
Sis.
 A: Basic idea:
$$ \lim_{n\to\infty}\int_0^1 \frac{e^{\displaystyle x^{n}}}{1+x^2}\,\mathrm{d}x = \int_0^1 \lim_{n\to\infty}\frac{e^{\displaystyle x^{n}}}{1+x^2}\,\mathrm{d}x$$
$$= \int_0^1 \frac{\lim_{n\to\infty}e^{\displaystyle x^{n}}}{1+x^2}\,\mathrm{d}x$$
$$= \int_0^1 \frac{e^{\displaystyle \lim_{n\to\infty}x^{n}}}{1+x^2}\,\mathrm{d}x$$
$$= \int_0^1 \frac{e^{\displaystyle 0}}{1+x^2}\,\mathrm{d}x$$
$$= \int_0^1 \frac{1}{1+x^2}\,\mathrm{d}x$$
$$ = {\pi \over 4}$$
Every step but the first is immediate or follows from continuity of an appropriate function. The first one is pretty standard (but not for high school students) since the integrands decrease in $n$. One way to prove it rigorously involves breaking the domain of integration into $[0,1-\epsilon]$ and $[\epsilon,1]$ parts. The second piece gives an integral less than $\epsilon$ for all $n$, and the integrands in the second converge uniformly (I know, not high school level but still) to ${1 \over 1 + x^2}$.
A: I'm might be missing something, but the integral is bounded by:
$$\int_0^1\frac{e}{1+x^2}dx > \int_0^1\frac{e^{x^n}}{1+x^2}dx > \int_0^1\frac{1}{1+x^2}dx$$
I think this is enough to interchange the limit with the integral due to the Dominated convergence theorem, arriving at:
$$\int_0^1\frac{1}{1+x^2}dx = \pi/4$$
