looking for $ \lim_{x^2+y^2 \rightarrow \infty} xye^{-(x+y)^2}$ explained. I've two related questions. The first is a solution I want to verify if my reasoning is correct, and the second one I fail to solve.
First one:
$$ \lim_{x^2+y^2 \rightarrow \infty }  x y e^{-x^2-y^2}$$ My solution is:
$$\frac{xy}{e^{x^2+y^2}} \rightarrow 0$$ that is, a simple rewrite and then you can "see" it.
The second question should not have a limit, however I would have calculated it to 0 as above. How should I reason in this case?
$$ \lim_{x^2+y^2 \rightarrow \infty} xye^{-(x+y)^2}$$
 A: First Problem:
I would rewrite the first limit as
$$
\lim_{x^2+y^2\to\infty}\frac{xy}{x^2+y^2}(x^2+y^2)e^{-x^2-y^2}\tag{1}
$$
For $u\ge0$,
$$
ue^{-u}\le\frac{u}{1+u+u^2}\le\frac1u\tag{2}
$$
Setting $u=x^2+y^2$ in $(2)$ yields
$$
(x^2+y^2)e^{-x^2-y^2}\le\frac1{x^2+y^2}\tag{3}
$$
Furthermore, for all $(x,y)\ne(0,0)$,
$$
\left|\frac{xy}{x^2+y^2}\right|\le\frac12\tag{4}
$$
Thus, $(1)$, $(3)$, and $(4)$ show that
$$
\lim_{x^2+y^2\to\infty}xy\,e^{-x^2-y^2}=0\tag{5}
$$

Second Problem:
Consider the path $(x,y)=(t,t)$. Then
$$
\lim_{x^2+y^2\to\infty}xy\,e^{(x+y)^2}=\lim_{t\to\infty}t^2\,e^{-4t^2}=0\tag{6}
$$
Consider the path $(x,y)=(t,-t)$. Then
$$
\lim_{x^2+y^2\to\infty}xy\,e^{(x+y)^2}=\lim_{t\to\infty}-t^2\,e^0=-\infty\tag{7}
$$
A: Hint: Try to keep $x+y=0$ while $x^2+y^2\to\infty$. Compare with keeping $x=0$ instead.
A: As pointed out; it is better to use polar coordinates to exploit symmetry.
For second part, note 
$xye^{-(x+y)^2}=r^{2}\sin\theta\cos\theta e^{-r^{2}(\sin\theta+\cos\theta)^{2}}$
Now observe when $\theta=-\cfrac{\pi}{4}$ above expression approaches $-\infty$ and $r^2$ approaches $\infty$
And when, say $\theta=0$ the expression is identically zero
So you can conclude...?
