TITLE: Use Dominated Convergence Theorem to find limit of $\int n f(x) e^{-nx} \, dx$
QUESTION [6 upvotes]: I was working on an old practice qual and I came across this problem.  I have a solution, but it's rather...convoluted...and I feel like there should be a simple way of using the dominated convergence theorem to solve it.  Anyways, here's the problem:

Let $f: [0,\infty) \rightarrow \mathbb{R}$ be Lebesgue integrable and suppose that $\lim_{x\rightarrow 0} f(x) = 2016$.  Show that $$\lim_{n\rightarrow \infty} \int_0^\infty nf(x)e^{-nx} \, dx = 2016$$

If you change variables, you get that the integral is equal to $\int_0^\infty f(\frac{x}{n})e^{-x}\, dx$, and so the claim is easy to prove if you know that $f$ is bounded.  In particular, if $f$ is continuous with compact support, the problem is solved.  However, despite the fact that we can prove the claim for a dense subset of $L^1$, it doesn't seem to follow for all of $L^1$ (as far as I can tell).
I'll put my solution (without directly using the DCT) in the answers because it's pretty long.

REPLY [4 votes]: Here is detail on my comment, as requested: 
We assume $c=\int_0^{\infty} |f(x)|dx<\infty$. 
Fix $\epsilon>0$ and notice that 
$$\left| \int_{\epsilon}^{\infty} f(x)ne^{-nx}dx \right| \leq ne^{-n\epsilon}\int_{\epsilon}^{\infty} |f(x)|dx \leq ne^{-n\epsilon}c \rightarrow 0$$
where the limit is as $n\rightarrow\infty$.  (You could also use LDC to claim this, as in my first comment).  So then
$$ \int_0^{\infty} = \int_0^{\epsilon} + \int_{\epsilon}^{\infty} $$
and $\lim_{n\rightarrow\infty} \int_0^{\epsilon}$ can be bounded (in terms of $\epsilon$) again by direct "sandwich" type methods.