A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 4 years ago

I need to show the following limit exists. lim(x,y) -> (0,0) of xy(x-y)/(x^2 + y^2). I can probably plug in for y=0 and x=0, but how can I show the limit exists for all lines?

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Try using polar coordinates: \[x = r\cos(\theta)\]\[y = r\sin(\theta)\] Now your limit becomes: \[\lim_{(x,y) \rightarrow (0,0)} [x \times y \times (x-y)] \div [x^{2} + y^{2}] = \] \[\lim_{r \rightarrow 0} [r \cos(\theta)\times r \sin(\theta)\times (r \cos(\theta) - r \sin(\theta))]\div[r^2 \cos^2(\theta) + r^2 \sin^2(\theta)] \] \[\lim_{r \rightarrow 0} [r^2\cos(\theta)\sin(\theta)\times r(\cos(\theta) - \sin(\theta))]\div r^2\]Now, observe that it's a limit of "r" going to 0, so r can never be 0, therefore we can divide by r² the top and the bottom our ratio:\[\lim_{r \rightarrow 0} r \times \cos(\theta)\sin(\theta)\times(\cos(\theta) -\sin(\theta)) = \]\[\cos(\theta)\sin(\theta)\times(\cos(\theta) -\sin(\theta))\lim_{r \rightarrow 0}r = 0\]Meaning that, independently of the angle (the theta) you are approaching the origin, it will always go to zero. Hope it helped!

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.