How to prove that $x^p \cdot \mathrm{e}^{-x} = O(x^{-2})$ When $x$ tends to $+\infty$ and where "O" is the Big-O-Notation
We recall the definition of big-O. That is $f(x) = O(g(x))$, if there exists constants $C,N$ such that $|f(x)| \leq C|g(x)| , \forall x &gt; N$. For this problem, it suffices to consider cases.  
Case 1: $p \leq 0$. Here it suffices to consider a constant $N  \geq 1$, for which $e^{-x} \leq x^{-2}$, $\forall x &gt; N$, because $x^{p}e^{-x} \leq e^{-x} \leq x^{-2}$. But $e^{-x} \leq x^{-2}$, $\forall x &gt; 2$. Hence, $x^{p} e^{-x} \leq x^{-2}$, $\forall x &gt; 2$, when $p \leq 0$.  
Case 2: $p &gt; 0$. We want to find a constant $N$ for which $x^{p} e^{-x} \leq x^{p} x^{-(p+2)} = x^{-2}$. Then we can find an $N$ for which $e^{-x} \leq x^{-(p+2)}$, $\forall x &gt; N$. However, 
$$
e^{-x} \leq x^{-(p+2)} \iff \frac{x}{ln(x)} \geq (p+2)
$$
Choose $N = (p+2)^{4} \implies \frac{(p+2)^{4}}{4\cdot (p+2)ln(p+2)} \geq 1$, for $p &gt; 0$. Altogether using the definition of big-O, we have that for arbitrary $p \in \mathbb{R}$, let $C = 1$. Then $x^{p}e^{-x} \leq x^{-2}$, $\forall x &gt; N = \max\{(p+2)^{4}, 2\}$. Hence $x^{p}e^{-x} = O(x^{-2})$
We want to prove $x^pe^{-x}=O(x^{-2})\iff x^{p+2}e^{-x}=O(1)$


In other words we want to show that $|x^{p+2}e^{-x}|$ is bounded.

This is obviously true since $$\forall r\in\mathbb R,\lim\limits_{x\to+\infty}x^re^{-x}=0$$ 

Thus it is bounded for instance by $1$ provided $x\gg 1$ is sufficiently large.




To show that exponential kills any monomial, we can deduce it from $\dfrac{\ln(x)}x\to 0$ at infinity.


In particular it is bounded by $\frac 1{r+1}$ for $r&gt;0$ and $x\gg 1$ sufficiently large.

Thus $(r+1)\ln(x)\le x\iff x^{r+1}\le e^x\iff x^re^{-x}\le \frac 1x\to 0$ 

The case $r\le 0$ does not pose an issue since in this case $x^r\le 1$ for $x\ge 1$ and $x^re^{-x}\le e^{-x}\to 0$ is easy.




The last point about $\ln(x)$ growth can be proven like this:


For $x\ge 1$ then $\frac 1x\le 1$ and we deduce that $f(x)=x-\ln(x)\nearrow$ since $f'(x)\ge 0$.

In particular $x-\ln(x)\ge f(1)=1\ge 0$ so $\ln(x)\le x$

If $x\ge 1$ so is $\sqrt{x}$ and we have $\ln(\sqrt{x})=\frac 12\ln(x)\le\sqrt{x}$

After dividing by $x$ we get $\dfrac{\ln(x)}x\le \dfrac 2{\sqrt{x}}\to 0$ at infinity.