How to evaluate $\lim_{n \rightarrow \infty} n\int_{[0,1]}f(x)e^{-nx}dx$ Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous. How do I compute the following limits?
$\lim_{n \rightarrow \infty} \int_{[0,1]}f(x)e^{-nx}dx$
and
$\lim_{n \rightarrow \infty} n\int_{[0,1]}f(x)e^{-nx}dx$
So, i the second limit it seems that the fact $f$ is continuous at $0$ will be very important. I've started this problem by considering the special case where there exists $\delta > 0$ such $f$ is constant on $[0,\delta]$. I've been trying to work on this problem for awhile now and am starting to get tunnel vision, I would appreciate some fresh insights! Thanks!
 A: Let $u=nx$. Then, provided everything is well behaved ($f$ having no $n$ dependence for example) we can use dominated convergence 
$$\lim_{n\to\infty} \int_0^n f\left(\frac{u}{n}\right)e^{-u}du = f(0)\int_0^\infty e^{-u}du = f(0)$$
A: For the  first one:
$\frac{1}{e^{nx}} \leq 1,\forall n \in \Bbb{N}$  and since $f$ is continuous on a compact intervals then it is bounded by some $M>0$
So $|f(x)e^{-nx}| \leq M \in L^1[0,1]$
Thus by Dominated convergence we have that the limit is zero.
In the second do the substitution $x=t/n$ and exploit the continuity of $f$ along with Dominated convergence.
A: There are $\xi_n\in[0,\sqrt{n}]$ and $\nu_n\in[\sqrt{n},n]$, s.t.
\begin{align*}
n\int_0^1f(x)e^{-nx}\mathrm{d}x&=\int_0^nf\left(\frac{x}{n}\right)e^{-x}\mathrm{d}x\\
&=\int_0^{\sqrt{n}}f\left(\frac{x}{n}\right)e^{-x}\mathrm{d}x+
\int_{\sqrt{n}}^nf\left(\frac{x}{n}\right)e^{-x}\mathrm{d}x\\
&=f\left(\frac{\xi_n}{n}\right)\int_0^{\sqrt{n}}e^{-x}\mathrm{d}x+f\left(\frac{\nu_n}{n}\right)\int_{\sqrt{n}}^ne^{-x}\mathrm{d}x.
\end{align*}
In this form, the rest should be easy to calculate.
