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

Here is a start. Making the change of variables $y=tx$ yields

$$ \int_0^de^{-tx}g(x)dx = \frac{1}{t}\int_{0}^{td} g\left(\frac{y}{t}\right)e^{-y} dy. $$

I think you can advance now.
Note:

$$ \int_{0}^{\infty} e^{-y}dy = 1. $$

