# I have to show that the limit goes to zero..

I need to show that the $$\lim_{t\to\infty} t^m e^{{\alpha}t}=0$$ where $m \in \mathbb N_0$ $\Re (\alpha)<0$ and

$$\max_{0 \leq t< \infty}\left|t^m\cdot e^{{\alpha}t}\right|= \left(\frac{m}{-\Re(\alpha)}\right)^m\cdot e^{-m}$$ $m \in \mathbb N_0$, $\Re (\alpha)<0$ by using L'Hopital's rule.

How do I start? Thank you so much! Klara

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Note that it is enough to show that the absolute value of your expression converges to zero, and so we can dispense with complex number.

Write $\alpha=\beta+i\gamma$. Then $\beta<0$ and, for $t\geq0$, $$|t^me^{\alpha t}|=t^m |e^{\beta t}e^{i\gamma t}|=t^me^{\beta t}.$$ Now, $t^me^{\beta t}=t^m/e^{-\beta t}$. As both numerator and denominator go to $\infty$, we apply L'Hôpital $m$ times to get $$\lim_{t\to\infty}\frac{t^m}{e^{-\beta t}}=\lim_{t\to\infty}\frac{mt^{m-1}}{-\beta e^{-\beta t}}=\ldots=\lim_{t\to\infty}\frac{m!}{(-1)^m\beta^me^{-\beta t}}=0,$$ as $-\beta>0$.

Now the function $t^me^{\beta t}$ is differentiable and goes to zero at infinity, so it achieves its maximum, and we can find it as a zero of its derivative. The derivative is $$mt^{m-1}e^{\beta t}+t^{m}\beta e^{\beta t}.$$ Equated to zero, we cancel $e^{\beta t}$ to get $$mt^{m-1}+\beta t^{m}=0,$$ so $t=-m/\beta$. Being the only zero of the derivative, it has to be a maximum. The value is $$\left(-\frac{m}\beta\right)^m\,e^{-\beta m/\beta}=\left(\frac{m}{-\beta}\right)^me^{-m},$$ i.e. $$\max_{0 \leq t< \infty}\left|t^m\cdot e^{{\alpha}t}\right|= \left(\frac{m}{-\Re(\alpha)}\right)^m\cdot e^{-m}.$$

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@ Martin Thank you so much! – Klara Nov 25 '12 at 0:39
You are welcome! – Martin Argerami Nov 25 '12 at 1:09

Assuming $\,t\in\Bbb R\,\,,\,\alpha = x+iy\,\,,\,x,y\in\Bbb R\,$:

$$\left|t^me^{\alpha t}\right|=\left|t^me^{tx}\right|=\frac{t^m}{e^{tw}}\stackrel{\text{L'Hospital}}{\xrightarrow [t\to\infty]{} 0}$$

Putting $\,w=-x=-Re(z)>0\,$ and for $\,t>0\,$ , which can be assumed as we want $\,t\to\infty\,$

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