MHB When taking the limit at infinity, is this allowed?

[find_the_fun]

In this video of Laplace transforms the equation $$\lim_{t \to \infty}\frac{te^{-st}}{-s}$$ is said to be 0. I'm not sure I agree with the reasoning. It says it's because $$e^t$$ grows faster than $$t$$; can you treat infinity like that? For example could you say $$\lim_{x \to \infty}\frac{x}{x^2}=0$$? I thought it was undefined because obviously $$\frac{infty}{infty}$$ is undefined.

 

[MarkFL]

The form $$\frac{\infty}{\infty}$$ is indeterminate...to allay your concerns, apply L'Hôpital's Rule. :D

 

[find_the_fun]

The form $$\frac{\infty}{\infty}$$ is indeterminate...to allay your concerns, apply L'Hôpital's Rule. :D

Maybe the video was aluding to L'Hôpital's Rule when it said $$e^t$$ grows faster than $$t$$.

By the way, according to here $$\frac{\infty}{\infty}$$ is indeterminate.