The limit of xye^-(x+y)^2 when x^2+y^2 approach infinity

[Mahathepp]

I try to figure it out but I can't get the answer that I need and when I look upon the solution from the book I don't understand it at all. The answer is " no limit" and there is no explanation why. The question is

Determine the limit of

lim _{(x2+y2)- -> infinity} (xye^-(x+y)2

in this case I use polar coordinate which I get

lim _{r2 - -> infinity} ( r²cos(x)sin(X) / e^{r^2(1+sin(2x)} )

My idea is since there is (e^{r^2(1+sin(2x))} in denominator which is depening on angle (2x) but I am not sure if I understand correct. Can anyone here help me to figure it out? Thanks in advance.

Regard

 

[Orodruin]

If the limit existed, then the result would not depend on how r approaches infinity. As you can see, you will obtain different results depending on the polar coordinate (if you fix it) and therefore the limit is not well defined.

 

[mathman]

If x+y = 0, the limit is -∞. If x+y -> ∞, the limit = 0. Manipulation of x+y can lead to any limit.

 

[LAZYANGEL]

$\lim\limits_{x^2+y^2\to\infty} x y e^{-(x+y)^2}$

$\lim\limits_{r^2\to\infty} r^2 \cos(\theta) \sin(\theta) e^{-r^2 (1+\sin(2\theta))}$

$\lim\limits_{r^2\to\infty} \frac {r^2 \cos(\theta) \sin(\theta)} {e^{r^2 (1+\sin(2\theta))}}$

The limit is undefined when the $e^{r^2 (1+\sin (2\theta))}=e^{\infty \cdot 0}$ or when $1+\sin(2\theta)=0$ and that happens for $\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$

$\arctan (\frac{y}{x})=\frac{3\pi}{4}$ And that means $\frac{y}{x}=-1$ or along the line $y=-x$ which restricts the domain.