Showing $g(x) = -2\pi x e^{-\pi x^2} = O(1/x^2)$ for $x \rightarrow \pm \infty$

I'm trying to understand Landau's notation and wanted to prove that

$$g(x) = -2\pi x e^{-\pi x^2} = O\left(\frac{1}{x^2}\right) \qquad \qquad \text{ for } x \rightarrow \pm \infty.$$

In the $x \rightarrow + \infty$ case, I think I have to prove that there is an $M > 0$ and $N > 0$ such that for all $x > N$ it holds that

$$\bigg|-2 \pi x e^{-\pi x^2}\bigg| \leq \frac{M}{x^2}.$$

But I'm stuck here as I don't seem to be able to prove that $2 \pi x^3 e^{-\pi x^2} \leq M.$

Solution 1:

If $|x|>1$ then $$ e^{\pi x^2 } = 1 + \pi x^2 + \frac{{\pi ^2 x^4 }}{{2!}} + \cdots > \frac{{\pi ^2 x^4 }}{{2!}} > x^4>|x|^3, $$ i.e., $$ e^{ - \pi x^2 } < \frac{1}{|x|^3} $$ for $|x|>1$. Then $$ \left| { - 2\pi xe^{ - \pi x^2 } } \right| < 2\pi \frac{1}{x^2} $$ for $|x|>1$.

Solution 2:

Because $\lim_{x \to +\infty}|-2\pi x^3 \exp (-\pi x^2)| = 0,$ we know we can find that $N$ such that for any $x > N$ we have $|-2\pi x^3 \exp (-\pi x^2)| \leq M.$ This already proves it.

In general, when $$\lim_{x\to a} \frac{||f(x)||}{h(x)}$$ exists (with $h(x)$ strictly positive for all $x\neq a$ in a neighbourhoud of $a$) and it's finite, so $$0 \leq \lim_{x\to a} \frac{||f(x)||}{h(x)} < +\infty, $$you can say that $f(x) = O(h(x))$ for $x \to a.$ This can be helpful in many cases, and it can also be helpful to get some intuition for what it means for $f(x)$ to be $O(h(x))$ for $x\to a.$

In the case of small o, it's the same thing but the limit has to be $0$ (if it exists). In the case of $\Omega$ it's kind of the opposite of big O. It has to be $$0 < \lim_{x\to a} \frac{||f(x)||}{h(x)} \leq +\infty,$$ again if it exists and in the case of $\Theta$ we want that the limit is a strictly positive number that isn't infinite.