How can I prove that $\mathcal L(t)$ equals $s^{-2}$?

Question

I'm trying to prove that $$\mathcal{L}(t) = \frac{1}{s^2}$$ from the definition.

After using integration by parts, I got:

$$\mathcal{L}(t)=\lim_{t \to \infty}\left (\frac{-t}{s} * e^{-st}-\frac{1}{s^2} * e^{-st}\right) - \lim_{t \to 0} \left(\frac{-t}{s} * e^{-st}-\frac{1}{s^2} * e^{-st}\right).$$

I have simplified this subtraction into

$$\lim_{t \to \infty} \left(\frac{-t}{s} \cdot\exp(-st)\right)+\frac{1}{s^2}$$ for every $s>0$.

So how can I prove that limit without using L'hopital?

Answer 1

$e^{st} \geq \frac {s^{2}t^{2}} 2$ from the series expansion. Hence $te^{-st}=\frac t {e^{st}} \le\frac {2t} {s^{2}t^{2}} \to 0$.