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

[Addez123]

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..

 

[BvU]

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 ...

 

[Addez123]

Ah, my bad. I should've defined it as: $$r^2 = x^2 + y^2$$

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$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

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

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?

 

[PeroK]

Ah, my bad. I should've defined it as: $$r^2 = x^2 + y^2$$

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$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

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

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):
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$

 

[Delta2]

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.

 

[PeroK]

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.

 

[BvU]

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 ?

 

[Addez123]

I think what @PeroK did is what they expect us to do. It's familiar to previous exercises.
@BvU I don't see how I would've solved for x and y to get v without going in circles like:
$$v = atan( \frac {r \cdot sin(v)}{r\cdot cos(v)})$$

 

[PeroK]

I think what @PeroK did is what they expect us to do. It's familiar to previous exercises.
@BvU I don't see how I would've solved for x and y to get v without going in circles like:
$$v = atan( \frac {r \cdot sin(v)}{r\cdot cos(v)})$$

You would do the equivalent of what I did but in polar coordinates. $v = \pi/2$ equates to $x = 0$; and $v = \pi/4$ equates to $x = y$.