What do limits of functions of the form $te^t$ have to do with l'Hopital's rule? I have an improper function that I have to integrate from some number to infinity. Once integration is done, the function is of the form $te^t$.
What I'm wondering is what does this have to do with l'Hopital's rule?
From reading my book, I see the following:

We know that $e^t \to 0$ as $t \to -\infty$, and by l'Hopital's rule we have
$$\lim_{t\to-\infty} te^t = \lim_{t\to-\infty} \frac{t}{e^{-t}} = \lim_{t\to-\infty} \frac{1}{-e^{-t}} = \lim_{t\to-\infty} -e^t = 0.$$

I know what l'Hopital's rule is

The limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

I don't understand what this has to do with the explanation above. Can somebody help me understand this please?

The invocation of l'Hôpital's rule is
$$ \lim_{t\to-\infty} \frac{t}{e^{-t}} = \lim_{t\to-\infty} \frac{1}{-e^{-t}}. $$
