Show that $\lim_{n \rightarrow \infty} n \int_{0}^{1}e^{-nx}f(x)dx=f(0)$ when $f:[0,1] \rightarrow \mathbb{R}$ is continuous. Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function.
Show that $\lim_{n \rightarrow \infty} n \int_{0}^{1}e^{-nx}f(x)dx=f(0)$
I am making a claim, 
$\lim_{n \rightarrow \infty} n \int_{0}^{1}e^{-nx}f(0) dx=f(0).$ Is it really true?

Use integration by parts:
$$ \lim_{n\to\infty} n\left[\left.{\frac{f(x)e^{-nx}}{n}}\right|^0_1-\frac{1}{n}\int_{0}^{1}f'(x)e^{-nx}\text{d}x\right] $$
$$ = f(0) - \lim_{n\to\infty}f(1)e^{-n} - \int_{0}^{1}\lim_{n\to\infty}f'(x)e^{-nx}\text{d}x $$
$$ = f(0) $$
