In deriving the Laplace transform of $t$, I'm not being able to grasp why the limit term go to 0:
$$\begin{align} x(t)&=t\\ \\ \therefore\; X(s)&=\int_0^\infty t\cdot e^{-st}\;\mathrm{d}t\\ &=\left[\frac{-te^{-st}}{s}-\frac{e^{-st}}{s^2}\right]_0^\infty \quad \text{(using integ. by parts)}\\ &=\lim_{t \to \infty}\left(\frac{-te^{-st}}{s}\right)-\left(0-\frac{1}{s^2}\right) \end{align}$$ We know that $$t \stackrel{\large \mathfrak{L}} \Longrightarrow \frac{1}{s^2} $$ $$\mathrm{Hence}\; \lim_{t \to \infty}\left(\frac{-te^{-st}}{s}\right)=0$$ But I can't fathom why it's so.