Bounding a polynomial from above How can one bound $t^m e^{\alpha t}$ ($m\in\mathbb{N}, \alpha<0$) so that 
$$t^me^{\alpha t} \le C_\delta e^{-\delta t}$$
 for some $C,\delta > 0$ (and for all $t\ge 0$?) 
For $0<\delta<-\alpha$ it is clear that $e^{\alpha t} \le e^{-\delta t}$, but my question is how to show that there exists $M_\delta$ that bounds the polynomial growth $t^m$?
 A: As long as $\alpha+\delta\lt0$, we can find the maximum of
$$
t^me^{(\alpha+\delta)t}
$$
by finding the point where the derivative is $0$.
$$
\frac{\mathrm{d}}{\mathrm{d}t}t^me^{(\alpha+\delta)t}=\left(\frac mt+\alpha+\delta\right)t^me^{(\alpha+\delta)t}
$$
which is $0$ when $t_0=-\frac{m}{\alpha+\delta}\gt0$; we know this is a maximum, because the derivative is positive for smaller $t$ and negative for larger $t$. Thus, we can set
$$
C_\delta=t_0^me^{(\alpha+\delta)t_0}
$$
and then
$$
t^me^{\alpha t}\le C_\delta e^{-\delta t}
$$
A: You won't be able to find a constant that bounds the polynomial, since $t^m\to\infty$ as $t\to\infty$.  You need another exponential to "take up the slack".  Specifically, take any $\delta$ such that $0<\delta<-\alpha$.  Then $-\alpha-\delta>0$ and so
$$t^m<e^{(-\alpha-\delta)t}\tag{$*$}$$
for all sufficiently large $t$.  Suppose that $(*)$ is true for all $t\ge t_0$.  The function $t^me^{(\alpha+\delta)t}$ is continuous on the interval $[0,t_0]$ and is therefore bounded on that interval, say
$$t^me^{(\alpha+\delta)t}<M_\delta\ .$$
Let $C_\delta=\max(1,M_\delta)$.  Then
$$\eqalign{
  0\le t\le t_0\quad&\Rightarrow\quad t^me^{\alpha t}<M_\delta e^{-\delta t}
    \quad\Rightarrow\quad t^me^{\alpha t}<C_\delta e^{-\delta t}\cr
  t\ge t_0\quad&\Rightarrow\quad t^me^{\alpha t}<e^{-\delta t}
    \quad\Rightarrow\quad t^me^{\alpha t}<C_\delta e^{-\delta t}\ .\cr}$$
So your required inequality is true for all $t\ge0$.
