Limit of an integral question: $\lim \limits _{h \to \infty} h \int \limits _0 ^\infty e ^{-hx} f(x) \, d x = f(0)$ Let $f : [0, \infty) \to \Bbb R$ be bounded and continuous. Prove that $\lim \limits _{h \to \infty} h \int \limits _0 ^\infty e ^{-hx} f(x) \, d x = f(0)$.
Our intuition was to use l'Hospital's rule to try to find the limit, but it doesn't seem to work.
Thanks
 A: Split the integral in two parts: $\int \limits _0 ^1$ and $\int \limits _1 ^\infty$.
For the second one, using the fact that $|f| \le M$, we obtain: $h \big| \int \limits _1 ^\infty \Bbb e ^{-hx} f(x) \Bbb d x \big| \le h \int \limits _1 ^\infty \Bbb e ^{-hx} |f(x)| \Bbb d x \le h \int \limits _1 ^\infty \Bbb e ^{-hx} M \Bbb d x = h M \frac {\Bbb e ^{-hx}} {-h} \big| _1 ^\infty = M \Bbb e ^{-h}$, which tends to $0$.
For the first one, we shall make the change of variable $y = hx$, thus obtaining: $h \int \limits _0 ^1 \Bbb e ^{-hx} f(x) \Bbb d x = \int \limits _0 ^h \Bbb e ^{-y} f(\frac y h) \Bbb d y$ which tends to $\int \limits _0 ^\infty \Bbb e ^{-y} f(0) \Bbb d y = f(0)$.
A: Because $h\int_{0}^{\infty}e^{-hx}dx=1$, you can write
$$
  \frac{1}{h}\int_{0}^{\infty}e^{-hx}f(x)dx-f(0)=\frac{1}{h}\int_{0}^{\infty}e^{-hx}\{f(x)-f(0)\}dx
$$
If $M$ is a bound for $f$ on $[0,\infty)$ and $r > 0$, then, as $h\rightarrow\infty$,
$$
          \left|h\int_{r}^{\infty}e^{-hx}\{f(x)-f(0)\}dx\right|
      \le 2Mh\int_{r}^{\infty}e^{-hx}dx=2Mhe^{-h\delta}\rightarrow 0
$$
Let $\epsilon > 0$ be given. Because $f$ is continuous at $0$, there exists $\delta > 0$ such that $|f(x)-f(0)| < \epsilon/2$ whenever $0 \le x <\delta$. And, there exists $h_{0}$ such that
$$
       \left|h\int_{\delta/2}^{\infty}e^{-hx}\{f(x)-f(0)\}dx\right| < \epsilon/2
   \mbox{ whenever } h > h_{0}.
$$
Putting the pieces together: If $h > h_{0}$, then
$$
       \left|\frac{1}{h}\int_{0}^{\infty}e^{-hx}f(x)dx-f(0)\right| < \epsilon.
$$
By the definition of limit,
$$
      \lim_{h\rightarrow\infty}h\int_{0}^{\infty}e^{-hx}f(x)dx =f(0).
$$
A: You may try to rewrite first and then take the limit
$$
h\int_0^\infty e^{-hx}f(x)\,dx-f(0)=\int_0^\infty e^{-t}\left[f\left(\frac{t}{h}\right)-f(0)\right]\,dt.
$$
