Maximization of $x^2e^{-(x^4+y^2)}$ My textbook proposes to maximize $f(x, y)=x^2e^{-(x^4+y^2)}$. It begins by showing that it tends to $0$ as $r=\sqrt{x^2+y^2}$ tends to $\infty$.
First it remarks that $x^2e^{-x}\rightarrow 0$ as $x\rightarrow\infty$, and that $e^{-x^4}$ is eventually smaller than $e^{-x}$, therefore $x^2e^{-x^4}$ tends to $0$.
Next it points out that as $y^2\geq0$, we have $e^{-y^2}\leq1$. From this I can see we can conclude that $f(x, y)=x^2e^{-(x^4+y^2)}=x^2e^{-x^4}e^{-y^2}\leq x^2e^{-x^4}$
From this it immediately concludes that

$f(x, y)\rightarrow 0$ as $r\rightarrow\infty$ Why?

This is my first question. While I can sort of see the justification for this, how do you prove it rigorously? I can see that, for instance, with fixed $y$, the limit is verified (because then we're just letting $|x|$ tend to infinity). Inversely, with fixed $x$, I suppose the limit would be verified as well (but I don't think that follows from anything the book explicitly points out - I think we need that $e^{-y}\rightarrow0$). Even given that, how do we know it'll still tend to $0$ if $(x, y)$ becomes large along some other axis?  I thought maybe I'd express $(x, y)$ as $tX$, where $t\geq0$ and $X$ is some vector, then letting $t\rightarrow\infty$. This results in an expression which I'm pretty sure does tend to $0$ (haven't worked through it in detail yet). But does that prove the desired proposition? And is there a simpler way of doing it that I'm missing (surely there must be or the book would have explained it in more detail)?
My second question is about what comes right after:

Hence any maximum occurs in a bounded region of the plane.

I'm actually having some difficulty understanding what this even means. That there exists a bounded region such that we can be sure no maximum occurs outside of it? If so, how? If $f$ does tend to $0$, then it's true that there's a bounded disc beyond which $f$ is smaller than, say, $1$. But maybe inside that region, $f$ is even smaller. Intuitively, I suppose that might mean there's a maximum on the boundary of that disc... not sure, though. And again, is there a simpler method that I'm missing?
 A: To see that $x^2e^{-(x^4+y^2)}\to0$ as $(x,y)\to0$, note that $e^{-(x^4+y^2)}\le1$ and $x^2\to0$ as $(x,y)\to0$.

For $x\ge3$, we have that
$$
\frac{(x+1)e^{-(x+1)}}{xe^{-x}}=\frac{x+1}{xe}\le\frac12
$$
Therefore, by induction, we have
$$
\frac{(x+n)e^{-(x+n)}}{xe^{-x}}\le\frac1{2^n}
$$
Thus, for $x\ge3$
$$
(x+n)e^{-(x+n)}\le\frac1{2^n}xe^{-x}
$$
which shows that
$$
\lim_{x\to\infty}xe^{-x}=0
$$
Substitute $x\mapsto2x^4$, divide by $2$, and take the square root to get
$$
\lim_{|x|\to\infty}x^2e^{-x^4}=0
$$
We also have
$$
\lim_{|y|\to\infty}e^{-y^2}=0
$$
Therefore, since $|x|\ge\frac{r}{\sqrt2}$ or $|y|\ge\frac{r}{\sqrt2}$
$$
\lim_{x^2+y^2\to\infty}x^2e^{-(x^4+y^2)}=0
$$
This says that
$$
\color{#C00000}{\lim_{r\to\infty}x^2e^{-(x^4+y^2)}=0}
$$

Note that
$$
x^2e^{-(x^4+y^2)}=x^2e^{-x^4}e^{-y^2}
$$
$e^{-y^2}\le1$ and is $1$ only when $y=0$.
For $x^2e^{-x^4}$, we can just take the derivative:
$$
\frac{\mathrm{d}}{\mathrm{d}x}x^2e^{-x^4}=(2x-4x^5)e^{-x^4}
$$
This vanishes precisely when $x^4=\frac12$ and that is at $\pm\frac1{\sqrt[4]{2}}$. At either of those points,
$$
x^2e^{-x^4}=\frac1{\sqrt2}e^{-1/2}
$$
Since $x^2e^{-x^4}$ tends to $0$ as $x\to0$ and $|x|\to\infty$, this must be the maximum.
