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?

Since $e^t=\frac{1}{e^{-t}}$, We have $$ \lim_{t\to-\infty}te^t=\lim_{t\to-\infty} \frac{t}{e^{-t}}$$
By the l'Hôpital's rule, we go on 
$$ = \lim_{t\to-\infty} \frac{1}{-e^{-t}}. $$
By the equation $e^t=\frac{1}{e^{-t}}$ again, we have 
$$ \lim_{t\to-\infty} \frac{1}{-e^{-t}}=\lim_{t\to-\infty}-e^t=0.$$
