Show that $\int_0^de^{-tx}g(x)dx\sim_{t\rightarrow\infty} \frac{g(0)}{t}$ Let $d>0$. Let $g\in C^0([0,d])$  such that $g(0)\ne0$.
Show that $$\int_0^de^{-tx}g(x)dx\sim_{t\rightarrow\infty} \frac{g(0)}{t}$$
How can I prove that ? 
It's the first time I see this kind of exercise
Thank you in advance for your time

Actually this limit is zero.
Proof.
Let's use the definition of limit:
$\forall\epsilon , \exists\delta \ |\int_0^de^{-tx}g(x)dx|<\delta$
As $g\in C^0([0,d])$, $|g|<c$, is bounded
In addition, it's known that $\lim_{t\rightarrow\infty}e^{-tx}=0$. Therefore for above $\epsilon ,\exists\delta_1 $ such $ |e^{-tx}|<\delta_1$ 
Putting all together:
$$\left|\int_0^de^{-tx}g(x)dx\right|< c \left|\int_0^de^{-tx}dx\right| < c \ d \ \delta_1 = \delta$$
This should be all.
