TITLE: What do limits of functions of the form $te^t$ have to do with l'Hopital's rule?
QUESTION [3 upvotes]: 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?

REPLY [2 votes]: Notice we're trying to evaluate the limit of a product which yields an indeterminant form:
$$\lim\limits_{t \to -\infty} te^t.$$
Considering the limit as is yields $-\infty\cdot0$. To avoid this, we utilize l'Hopital's rule.
Now to apply l'Hopital's rule, we need to write $te^t$ as a quotient, i.e., $\dfrac{t}{e^{-t}}$. We now can use l'Hopital's rule to evaluate the limit.
$$\lim\limits_{t \to -\infty} te^t = \lim\limits_{t \to -\infty} \dfrac{t}{e^{-t}} \underbrace{= \lim\limits_{t \to - \infty} \dfrac{(t)'}{(e^{-t})'}}_{\mbox{by l'Hopital}} = \lim\limits_{t \to - \infty} \frac{1}{-e^{-t}} = \lim\limits_{t \to - \infty} -e^{t} = 0.$$
Hence, we see $\lim\limits_{t \to - \infty} te^t = 0$.
So, regarding the improper integral you are solving: can you apply a similar trick to evaluate the limit you're left with?

REPLY [0 votes]: 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 
$$ \lim_{t\to-\infty} \frac{1}{-e^{-t}}=\lim_{t\to-\infty}-e^t=0.$$
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