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?
 A: Without using LDCT 
We have
$$I_n = \int_0^1 ne^{-nx}f(x)\, dx= \int_0^1 ne^{-nx}[f(x) - f(0)] \, dx + f(0)\int_0^1 ne^{-nx} \, dx \\ = \int_0^c ne^{-nx}[f(x) - f(0)] \, dx +  \int_c^1 ne^{-nx}[f(x) - f(0)] \, dx + f(0)(1- e^{-n}),$$
and
$$\left| I_n - f(0)\right| \leqslant \int_0^c ne^{-nx}|f(x) - f(0)| \, dx +  \int_c^1 ne^{-nx}|f(x) - f(0)| \, dx + |f(0)|e^{-n}$$
For any $\epsilon > 0,$ choose $c$ sufficiently small such that $|f(x) - f(0)| < \epsilon$ for $ 0 < x < c$.
Then
$$|I_n - f(0)| \leqslant \epsilon(1 - e^{-nc}) + (2\sup_{x \in [0,1]} |f(x)|)(e^{-nc}- e^{-n}) + |f(0)|e^{-n} .$$
Taking the limit as $n \to \infty$ we get
$$\lim_{n \to \infty}|I_n - f(0)| \leqslant \epsilon.$$
Since $\epsilon$ is arbitrarily small, it follows that $\lim I_n = f(0).$
Using LDCT
Alternatively, changing variables with $u = nx,$ we have
$$n \int_{0}^{1}e^{-nx}f(x)dx=\int_0^ne^{-u}f(u/n) \, du = \int_0^{\infty}e^{-u}f(u/n)1\{u \leqslant n\} \, du $$
Since $f$ is continuous, $f(u/n) \to f(0)$ as $n \to \infty$, and
$$\lim_{n \to \infty}e^{-u}f(u/n)1\{u \leqslant n\} = f(0)e^{-u}$$
By the Lebesgue dominated convergence theorem
$$\lim_{n \to \infty}\int_0^{\infty}e^{-u}f(u/n)1\{u \leqslant n\} \, du = \int_0^{\infty}f(0)e^{-u}\, du = f(0)$$
A: Use integration by parts:
$$ \lim_{n\to\infty} n\left[\left.{\frac{f(x)e^{-nx}}{n}}\right|^0_1-\frac{1}{n}\int_{0}^{1}f'(x)e^{-nx}\text{d}x\right] $$
$$ = f(0) - \lim_{n\to\infty}f(1)e^{-n} - \int_{0}^{1}\lim_{n\to\infty}f'(x)e^{-nx}\text{d}x $$
$$ = f(0) $$
