Why does $\lim\limits_{t\to-\infty}te^t=0$?
Solution 1:
We can definitely use L'Hôpital's rule: $$\lim_{t\to-\infty}te^t=\lim_{t\to-\infty}\dfrac{t}{e^{-t}}=\lim_{t\to-\infty}\dfrac{1}{-e^{-t}}=0.$$
Solution 2:
Set $x=-t$
$$\lim_{t\to -\infty}te^t=-\lim_{x\to +\infty}\frac{x}{e^x}=0$$
L'Hôpital's rule is not necessary
Solution 3:
You can avoid L'hopital's rule by noting by definition, $e^x = \sum_n \frac{x^n}{n!}$.
Consider $\lim_{x \to \infty} x e^{-x} = \lim_{x \to \infty} \frac{x}{e^x}$.
Now, note for $x>0$ by definition of $e^x$, since all terms in the sum are non-negative, $e^x \geq 1+x+(1/2) x^2$ so $\frac{1}{1+x+(1/2)x^2} \geq \frac{1}{e^x}$. Multiplying both sides by $x$ preserves the inequality for $x>0$.
Clearly, since $\frac{x}{e^x} \geq 0$, $\lim_{x \to \infty} \frac{x}{e^x} \geq 0$ and $\frac{x}{1+x+(1/2)x^2} \geq \frac{x}{e^x}$ implies $\lim_{x \to \infty} \frac{x}{1+x+(1/2)x^2} \geq \lim_{x \to \infty} \frac{x}{e^x} $. But the limit on the left hand side is $0$ (its linear/quadratic), so we have $\lim_{x \to \infty} \frac{x}{e^x} = 0$.