Solve limits in Lebesgue integral Solve the limits of below:
(1) $\lim\limits_{n \to  \infty} \int_0^n (1+\frac{x}{n})^n e^{-2x}dx$.
(2) $\lim\limits_{n \to \infty} \int_0^n (1-\frac{x}{n})^n e^{\frac{x}{2}}dx$. 
(3) $\int_0^{\infty} \frac{e^{-x}\sin^2 x}{x}dx$
(Hint:  show $f(x,y)=e^{-x}\sin2xy$ is integrable on $[0,\infty) \times [0,1]$)
 A: For first and second questions, use $1-a\le e^{-a}, 1+a\le e^a $ for $0\le a\le 1$. (To prove, use Taylor remainder theorem.) Then use Lebesgue dominated convergence theorem.  For the third question, use Lebesgue dominated convergence theorem, note that $\sin x/x$ is bounded.
Also note that limit of $(1\pm x/n)^n$ is $e^{\pm x}$ as $n$ approaches to infinity.
A: In (2), define a function $f_n$ to be $(1-\frac{x}{n})^ne^{x/2}$ on $[0,n]$ and 0 for $x>n$. Then $f_n \rightarrow e^{-x/2}$ as $n\rightarrow \infty$, and furthermore, $e^{-x/2} \ge |f_n(x)|$ for all $n$, and $e^{-x/2} \in \mathcal{L}^1(R^1)$. 
So by Lebesgue's dominated convergence theorem,
$ \lim\limits_{n\to \infty} \int_0^{\infty} f_ndx= \int_0^{\infty} e^{-x/2}dx=2 $
and  the left-hand side is the first limit to compute. 
By similar considerations, in (1), $f_n \mapsto e^{-x}$ and $\int_R e^{-x}dx <\infty$ (i.e. $e^{-x}\in \mathcal{L}^1(R)$),
then the integral is 
$\int_0^{\infty} e^{-x}dx=1$
As for (3), $\int_0^1 e^{-x}sin2xydy=-\frac{e^{-x}cos2xy}{2x}|_0^1=\frac{(1-cos2x)e^{-x}}{2x}=\frac{e^{-x}sin^2x}{x}$.
