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

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In summary: 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##.
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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..
 
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  • #2
Addez123 said:
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 ...
 
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  • #3
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?
 
  • #4
Addez123 said:
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):
$$\lim_{x^2+y^2 \rightarrow +\infty} {\frac {xy} {x^2+y^2}}$$
 
  • #5
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.
 
  • #6
Delta2 said:
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.
 
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  • #7
Addez123 said:
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 ?
 
  • #8
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:
$$v = atan( \frac {r \cdot sin(v)}{r\cdot cos(v)})$$
 
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  • #9
Addez123 said:
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##.
 

1. What does "limit when x^2 + y^2 -> inf" mean?

This means that we are taking the limit of a mathematical expression, specifically when the sum of the squares of x and y approaches infinity. In other words, we are looking at what happens to the expression as x and y get larger and larger.

2. How do I solve for this limit?

To solve for this limit, we can use various mathematical methods such as substitution, algebraic manipulation, and trigonometric identities. It is important to carefully analyze the expression and use the appropriate techniques to arrive at the correct solution.

3. What is the significance of taking the limit when x^2 + y^2 -> inf?

This limit is often used in calculus and other areas of mathematics to analyze the behavior of a function as the input values approach infinity. It can help us understand the overall trend or pattern of the function and its behavior at extreme values.

4. Are there any common mistakes when solving this type of limit?

Yes, there are a few common mistakes that can occur when solving this type of limit. These include incorrectly applying mathematical rules, not considering all possible cases, and making errors in algebraic manipulation. It is important to check your work and carefully analyze the problem to avoid these mistakes.

5. Can this limit have more than one solution?

Yes, it is possible for this limit to have multiple solutions. This can occur when the expression has different behaviors at different values of x and y. It is important to carefully analyze the limit to determine if there are any restrictions or conditions that may affect the solution.

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