# Limit when x^2 + y^2 -> inf, am I solving it correctly?

Homework Statement:
Decide if the limit exists.
Relevant Equations:
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$
I'm not sure if the way I solve these limits is correct, so let me know if I'm doing something wrong.
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$
$$r = x^2+y^2$$
$$\lim_{r \rightarrow +\infty} {\frac {r\cdot cos(v) \cdot r \cdot sin(v)} r}$$
$$\lim_{r \rightarrow +\infty} {r\cdot cos(v)sin(v)} \rightarrow \infty$$

Also how can I be sure if $$cos(v)\cdot sin(v)$$ will make this go to inf, -inf or even be undefined?
I havn't even defined v..

Homework Helper
I havn't even defined v..

Not only that, you define ##r = x^2+y^2## which means ##x \ne r\cos\theta##
Try again, now with ##x =\sqrt r\cos\theta##

And think about what the role and meaning of ##\theta## could be ...

HallsofIvy
Ah, my bad. I should've defined it as: $$r^2 = x^2 + y^2$$
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

$$\lim_{r^2 \rightarrow +\infty} {\frac {r\cdot cos(v) \cdot r \cdot sin(v)} {r^2}}$$
$$\lim_{r^2 \rightarrow +\infty} {cos(v)sin(v)}$$

Well, this just made it trickier. I have no idea what the angle is.
v = atan(y/x) doesn't really help..

Should I have solved this in a completely different way?

Homework Helper
Gold Member
2022 Award
Ah, my bad. I should've defined it as: $$r^2 = x^2 + y^2$$
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

$$\lim_{r^2 \rightarrow +\infty} {\frac {r\cdot cos(v) \cdot r \cdot sin(v)} {r^2}}$$
$$\lim_{r^2 \rightarrow +\infty} {cos(v)sin(v)}$$

Well, this just made it trickier. I have no idea what the angle is.
v = atan(y/x) doesn't really help..

Should I have solved this in a completely different way?

Whenever you are stuck like this you should ask: what is the definition of the thing I'm trying to prove? In this case, what does this mean (in precise mathematical terms):
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

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Gold Member
I think your work (after the correction regarding r) is good. You find that the limit is ##\cos v\cdot\sin v## which implies that it depends on ##v## which implies that the limit does not exist.

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Gold Member
2022 Award
I think your work (after the correction regarding r) is good. You find that the limit is ##\cos v\cdot\sin v## which implies that it depends on ##v## which implies that the limit does not exist.

Alternatively, of course, you could choose a sequence of points with ##x = 0## and ## y \rightarrow \infty##, where the limit for that sequence is ##0##. Then, a sequence of points with ##x =y##, where the limit is ##1/2##.

From these examples alone you can see the limit does not exist.

Homework Helper
Well, this just made it trickier. I have no idea what the angle is.
v = atan(y/x) doesn't really help..
But it should ! ##\ ## I had silently hoped that this would have reminded you of another non-existing limit, $$\lim_{(x,y) \downarrow (0,0)}\; {y\over x}$$ where ##{y\over x}## similarly has a value that depends on the angle at which you approach the origin...

Now that you have been given the answer, is it clear ?

$$v = atan( \frac {r \cdot sin(v)}{r\cdot cos(v)})$$
$$v = atan( \frac {r \cdot sin(v)}{r\cdot cos(v)})$$