Laplace Transform

Question

Suppose that $F(s)=L[f(t)]$ and $G(s)=L[g(t)]$, where $L$ is the Laplace transformation $$F(s)=L[f(t)]=\int_0^{+\infty}e^{-st}f(t)dt.$$ I'm trying to prove that: $$\textrm{If}\ \ \lim_{t\to 0^+} \frac{f(t)}{g(t)}=1,\ \ \ \textrm{then}\ \ \ \lim_{s\to+\infty}\frac{F(s)}{G(s)}=1.$$ Can someone give me a hint? Thanks.

Answer 1

Hint :

$$\lim_{t\to 0} f(t) = \lim_{s\to \infty}sF(s)$$

This is the Initial value theorem.

---

Edit:

Expanding on this, since the limit we're seeking for is known to exist from the hypothesis, you may proceed as such :

$$\lim_{t\to 0} \frac{f(t)}{g(t)} = \frac{\lim_{t\to 0} f(t)}{\lim_{t\to 0}g(t)}= \frac{\lim_{s\to \infty}sF(s)}{\lim_{s\to \infty}sG(s)} = \lim_{s\to \infty}\frac{F(s)}{G(s)}$$

But:

$$ \lim_{t\to 0} \frac{f(t)}{g(t)} = 1$$

This is basically it.