When taking the limit at infinity, is this allowed?

[find_the_fun]

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

 

[MarkFL]

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

 

[find_the_fun]

The form \(\displaystyle \frac{\infty}{\infty}\) 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 \(\displaystyle e^t\) grows faster than \(\displaystyle t\).

By the way, according to here \(\displaystyle \frac{\infty}{\infty}\) is indeterminate.