More concretely, I have the integral
$$\lim_{n\to\infty}\int_{\left(0,\frac{n}{2}\right)}x^2e^x\left(1-\frac{2x}{n}\right)^nd\lambda(x)$$
It is clear that this is the same as
$$\lim_{n\to\infty}\int_{\left(0,\frac{n}{2}\right)}x^2e^xe^{-2x}d\lambda(x)$$
Which them simplifies to $$\lim_{n\to\infty}\int_{\left(0,\frac{n}{2}\right)}x^2e^{-x}d\lambda(x)$$ What are the exact rules that apply to such limits? Do I need to mention them?
On
Considering that the $n$ dependence is in the integral parameter, I think we can just use:
$\int_{(0,\frac{n}{2})}x^2e^x(1-\frac{2x}{n})^nd\lambda\leq\int_{(0,\frac{n}{2})}x^2e^{-x}d\lambda(x)$ Now try integration by parts and then take the limit.
Integration by parts gives us $\int_{(0,\frac{n}{2})}x^2e^{-x}d\lambda(x)=-e^{\frac{-n}{2}}\frac{n^2}{4}-e^{\frac{-n}{2}}n-2e^{\frac{-n}{2}}+2\rightarrow 2$, as $n\rightarrow\infty$
The idea behind this integral is to apply the dominated convergence theorem for measurable functions. First of all define $f(x) := 1_{[0,\frac{n}{2}]}x^2e^x\left( 1-\frac{2x}{n}\right)^n$. We note that
$$\lim_{n\to \infty}f(x) = \lim_{n\to \infty} 1_{[0,\frac{n}{2}]}x^2e^x\left( 1-\frac{2x}{n}\right)^n = 1_{[0,\infty]}x^2e^xe^{-2x} = 1_{[0,\infty]}x^2e^{-x}=:f$$
$f$ is measurable since is a composition of measurable functions. Moreover
$$|f(x)| = |1_{[0,\frac{n}{2}]}x^2e^x\left( 1-\frac{2x}{n}\right)^n|\le 1_{[0,\infty]}x^2e^xe^{-2x}=x^2e^{-x}=:g(x)$$
and
$$\int g(x)d\lambda = \int1_{[0,\infty]}x^2e^{-x}d\lambda=\int_0^{\infty}x^2e^{-x}d\lambda=2\lt\infty$$
Now we are allowed to apply the dominated convergence theorem since all the conditions needed are satisfied and we get:
$$\lim_{n\to\infty}\int_{\left(0,\frac{n}{2}\right)}x^2e^x\left(1-\frac{2x}{n}\right)^nd\lambda(x)= \lim_{n\to\infty}\int_{-\infty}^{\infty}1_{[0,\frac{n}{2}]}x^2e^x\left(1-\frac{2x}{n}\right)^nd\lambda(x)$$
$$=\int_{-\infty}^{\infty}\lim_{n\to\infty}1_{[0,\frac{n}{2}]}x^2e^x\left(1-\frac{2x}{n}\right)^nd\lambda(x)=\int_0^{\infty}x^2e^{-x}d\lambda=2$$
Hope this helps.