# Show $\lim_{t \to \infty} -t^{x}e^{-t} = 0$ [duplicate]

I have this limit (derived from an improper integral in the larger problem) that WolframAlpha says equals zero. But it doesn't have the steps to get there. How indeed do I show this instead of waving my hands that we can apply l'Hospital's rule to infinity and the denominator will eventually win and the limit will approach zero.

Check form ($\infty * 0$): $$\lim_{t \to \infty} -t^{x}e^{-t}$$ Rearrange into $\frac\infty\infty$: $$= \lim_{t \to \infty} \frac{-t^x}{e^t}$$ Apply l'Hospital's Rule: $$= \lim_{t \to \infty} \frac{-xt^{x-1}}{e^t}$$ I don't see a clear way of showing this goes to zero, but zero is the correct answer.

• $x$ is a constant, correct? – Mark Sep 11 '17 at 0:21
• You didn't apply L'H correctly... – Simply Beautiful Art Sep 11 '17 at 0:22
• @Mark, yes, the parent integral is with respect to u. – Matt Sep 11 '17 at 0:22
• It is $-t^x=-e^{\ln(t)x}$ – Cornman Sep 11 '17 at 0:23
• @SimplyBeautifulArt, it's been 3 years since I've used calculus, what did I do wrong? – Matt Sep 11 '17 at 0:23

Hint: Try and use l'hopital $x$ times, and then show:
$$\lim_{t \to \infty} \frac{1}{e^t}=0 \implies \lim_{t \to \infty} \frac{t}{e^t}=0 \implies \lim_{t \to \infty} \frac{t^2}{e^t}=0 ...$$
Because differentiating $t^x$ leads to $t^{x-1}$ multiplied by a constant (and constants don't matter in limits).