Bound the tails of $\exp(- \|x\|^2 / (2\sigma^2))$

Question

Problem

Consider the (multivariate Gaussian density) function $$ f_\sigma(x) = \frac{1}{(2\pi)^{m/2}\sigma^m}\exp\left(-\frac{\|x\|^2}{2\sigma^2}\right). $$ I want to show that there exists a constant $A$ such that $$ |f_\sigma(x)| \leq \frac{A\sigma}{\|x\|^{m+1}} $$ for any $\sigma > 0$ and for any $x\in\mathbb{R}^m\backslash\{0\}$.

Attempted Solution

I know I can bound $\exp( -\|x\|^2 / (2\sigma^2) ) \leq 1$, which gives me $$ |f_\sigma(x)| \leq \frac{1}{(2\pi)^{m/2}\sigma^m}. $$ I could multiply and divide by $\sigma$ but then not sure how to proceed.

Answer 1

This follows from the fact that $$r^{m+1} e^{-r^2/2} \longrightarrow 0$$ as $r \rightarrow \infty$. So, this function is bounded by some constant. Replace $r$ by $\|x\|/\sigma$ to conclude.