# why is $\lim_{t \to \infty}t^m e^{-\alpha t} = 0$ for every $m \in \mathbb{N}$ fixed and $\alpha \in \mathbb{C}$ with $Re(\alpha) \gt 0$?

I'm having trouble trying proving this fact:

$\lim_{t \to \infty}t^m e^{-\alpha t} = 0$ for every $m \in \mathbb{N}$ fixed and $\alpha \in \mathbb{C}$ with $Re(\alpha) \gt 0$

I tried to use l'Hôpital's rule, but I'm not going anywhere...

$$\lim_{t \to \infty}t^m e^{-\alpha t}= \lim_{t \to \infty}\frac{e^{-\alpha t}}{\frac{1}{t^m}}=\lim_{t \to \infty}\frac{-\alpha e^{-\alpha t}}{\frac{-m}{t^{m+1}}}=\lim_{t \to \infty}\frac{\alpha ^2e^{-\alpha t}}{\frac{m(m+1)}{t^{m+2}}}=...=\lim_{t \to \infty}\frac{(-1)^n\alpha ^ne^{-\alpha t}}{\frac{(-1)^nm(m+1)\dot{}\dot{}\dot{}(m+n-1)}{t^{m+n}}}$$

$$=\lim_{t \to \infty}\frac{t^{m+n}(-1)^n\alpha^ne^{-\alpha t}}{(-1)^nm(m+1)\dot{}\dot{}\dot{}(m+n-1)}$$

EDIT Following the Hint that orion gave me I got:

$$\lim_{t \to \infty}t^m e^{-\alpha t}= \lim_{t \to \infty}\frac{t^m}{\frac{1}{e^{-\alpha t}}}$$

Since

$$\frac{\delta^m}{\delta t^m}(t^m)=m! \quad\text{and}\quad \frac{\delta^m}{\delta t^m}\left(\frac{1}{e^{-\alpha t}}\right)=\frac{\alpha^m}{e^{-\alpha t}}$$

we have that $$\lim_{t \to \infty}\frac{t^m}{\frac{1}{e^{-\alpha t}}}=\lim_{t \to \infty}\frac{m!e^{-\alpha t}}{\alpha ^m}=0 \quad\text{since m fixed}$$

## 2 Answers

First of all, you're doing l'Hospital in the wrong direction. Put $t^m$ in the numerator, so that derivative actually decreases the power. Then, $m^{\text{th}}$ derivative will give you the answer you seek.

You could just as well prove that $e^{\alpha t}$ increases quicker than any finite power $t^m$. You can see that from the comparison of Taylor series.

$\large \lim_{t\to \infty}\dfrac{t^m}{e^{\alpha t}}$ since the limit is in the inderteminate form $\frac{\infty}{\infty}$

you can use l'hopital

$\lim_{t\to \infty}\dfrac{t^m}{e^{\alpha t}}=\frac{m}{\alpha}\lim_{t\to \infty}\dfrac{t^{m-1}}{e^{\alpha t}}=\frac{m(m-1)}{\alpha^2}\lim_{t\to \infty}\dfrac{t^{m-2}}{e^{\alpha t}}=\frac{m(m-1)(m-2)}{\alpha^3}\lim_{t\to \infty}\dfrac{t^{m-3}}{e^{\alpha t}}=...\\=\frac{m!}{\alpha^m}\lim_{t\to \infty}\dfrac{1}{e^{\alpha t}}=\frac{m!}{\alpha^m}*0=0$

since $Re(\alpha)>0$ this is valid