Showing that $x^me^{-ax}$ is bounded Let $a>0$, $m\geq 0$ and let $f:[0,\infty)\rightarrow\mathbb{R}$ be defined as $f(x)=x^me^{-ax}$. It should be true that this function is bounded, because near $0$ both terms are bounded, and far from $0$ the term $e^{-ax}$ decreases more rapidly than the term $x^m$ increases. What is a nice way to show this rigorously?

Since
$$
\lim_{x\to \infty}f(x)=0,
$$
there exists $r>0$ such that 
$$
f(x) \le 1 \quad \forall x \ge r.
$$
Thus for every $x \ge 0$ we have
$$
f(x) \le \max\{\max_{0 \le x \le r}f(x),1\}.
$$
