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 $\lim_{x \to \infty} \frac{x}{e^x} = 0$.