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Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Using polar coordinates, show that lim (x,y)->(0,0) [sin(x^2+y^2)]/[x^2+y^2] = 1

    2. Relevant equations

    r^2=x^2+y^2

    3. The attempt at a solution

    I was able to get the limit into polar coordinates:

    lim r->0^+ [sin(r^2)]/r^2

    but I'm not sure how to take this limit. I tried L'Hospital's, but it was just messy and useless seeming. So, any hints?
     
  2. jcsd
  3. Oct 6, 2012 #2

    jbunniii

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    What went wrong when you tried to use L'Hospital's rule? Where did you get stuck?
     
  4. Oct 6, 2012 #3
    I got:

    lim r->0^+ [sin(r^2)(2r)-r^2cos(r^2)(2r)]/r^4

    which simplifies to:

    lim r->0^+ [2sin(r^2)-2r^2cos(r^2)]/r^3

    but that's still 0/0 form, so it got me nowhere useful as far as I can see.
     
  5. Oct 6, 2012 #4

    SammyS

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    Applying L'Hôpital's rule gives you:
    [itex]\displaystyle \lim_{r\,\to\,0^{+}}\frac{\sin(r^2)}{r^2}[/itex]

    [itex]\displaystyle =\lim_{r\,\to\,0^{+}}\frac{d(\sin(r^2))/dr}{d(r^2)/dr}[/itex]​
    This is not what you have.
     
    Last edited: Oct 7, 2012
  6. Oct 6, 2012 #5

    LCKurtz

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    If you let ##\theta=r^2## does the limit become anything you have already studied?
     
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