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Let $P$ and $Q$ be two arbitrary polynomials. Why do we have the following

\begin{equation} P(x)Q(x)e^{-x^2} = o \Big(\frac{1}{x^2} \Big)? \end{equation}

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    Hint: each polynomial has a maximal degree. Let $M$ be the maximum of the two, so that your polynomials are bounded above by $Cx^M$ for some constant $C$. Alternatively, the product of two polynomials is another polynomial. Can you finish it from here? – 2017-01-12
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    If I bound above the product of $P$ and $Q$ by say $Mx^{\mathrm{deg}(P)+\mathrm{deg}(Q)+1}$, what is the argument that allows me to say that the exponential grows faster than this polynomial? Is it the fact that when I rewrite the exponential as a power series, then we obtain something of the sort: $\frac{1}{\sum_{n \geq 0} \frac{x^{n-\mathrm{deg}}}{n!}}$ which I may then bound by $\frac{1}{n^2}$? – 2017-01-12
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    The simplest argument would be using l'Hopital's rule. – 2017-01-12
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    You can simply use the fact that $x^{n} e^{-x} \to 0$ as $x\to\infty$ for all positive real numbers $n$. Your question is asking you to prove that $f(x)e^{-x^{2}}\to 0$ where $f(x) =x^{2}P(x)Q(x)$ is a polynomial. – 2017-01-12

2 Answers 2

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Let $R(x)=P(x)Q(x)$ be a third polynomial so that our problem reduces down to

$$R(x)e^{-x^2}=o(x^{-2})$$

One should then see that

$$|R(x)|e^{-x^2}<\epsilon x^{-2}$$

or,

$$e^{-x^2}<\frac\epsilon{x^2R(x)}$$

or,

$$e^{x^2}>x^{3+M}>\frac{x^2R(x)}\epsilon$$

where $M$ is the highest degree of $R(x)$.

Which should then make it more obvious.

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    I know that my question may be a bit obvious, but I mainly wanted to know what allows me to say that $e^{-x^2} \leq x^{-2}$. – 2017-01-12
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    @bgsk If we flip the inequality, we end up with$$e^{x^2}\ge x^2$$Or similarly,$$e^x>x$$which can be proven through various methods – 2017-01-12
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Let $d_1$ be the degree of $P(x)$ and $d_2$ be the degree of $Q(x)$. If we choose $n\in\mathbb{N}$ such that $2n\geq d_1+d_2+3$, by exploiting $\exp(z^2)\geq 1+z^2$ we have:

$$ P(x)Q(x) e^{-x^2} = \frac{P(x)Q(x)}{\exp(x^2/n)^n} = O\left(\frac{x^{d_1+d_2}}{x^{2n}}\right) = O\left(\frac{1}{x^3}\right) = o\left(\frac{1}{x^2}\right) $$ as $x\to +\infty$.