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

1. Mar 18, 2015

### 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) / er^2(1+sin(2x) )

My idea is since there is (er^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

Last edited: Mar 18, 2015
2. Mar 18, 2015

### Orodruin

Staff Emeritus
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.

3. Mar 18, 2015

### mathman

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

4. Mar 21, 2015

### 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.