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Multivariable limit (x,y) to (0,0)

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the limit:

    [itex]lim_{(x,y)\rightarrow(0,0)}=\frac{4\sqrt{x^{14}+y^{14}}}{(x^{2}+y^{2})^{3}}[/itex]

    3. The attempt at a solution

    I think I'm supposed to use algebra for this but I can't seem to figure anything useful out.

    I've found this:

    [itex]x^{14}+y^{14}=(x^{2}+y^{2}) (x^{12}-x^{10} y^{2}+x^{8} y^{4}-x^{6} y^{6}+x^{4} y^{8}-x^{2} y^{10}+y^{12})[/itex]

    but I'm not sure that helps me at all.

    I know the limit exists and it's zero because I've plotted it.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 24, 2012 #2

    SammyS

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    Change this to polar coordinates.
     
  4. Jan 24, 2012 #3
    Thank you for the advice.

    I plugged in [itex]x=rcos(\theta)[/itex] and [itex]y=rsin(\theta)[/itex] and after a little bit of cleaning up I got:

    [itex]lim_{(r,\theta)\rightarrow(0,0)}=4r\sqrt{cos^{14}(\theta)+cos^{14}(\theta)}[/itex]

    But the limit is supposed to be one... I'm note sure I'm doing this right, or maybe I just don't understand this as well as I wanted. :blushing:
     
  5. Jan 24, 2012 #4

    SammyS

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    That should be [itex]\displaystyle\lim_{r\to 0}\ 4r^5\sqrt{\cos^{14}(\theta)+\sin^{14}(\theta)}\,.[/itex]

    θ does not go to zero.

    The limit is zero, not 1.
     
  6. Jan 24, 2012 #5
    Yep of course the limit is zero, sorry about that.

    This is how I calculated my result:

    [itex]lim_{(x,y)\rightarrow(0,0)}\frac{4\sqrt{x^{14}+y^{14}}}{(x^{2}+y^{2})^{3}}=lim_{r\rightarrow0}\frac{4\sqrt{r^{14}cos^{14}(\theta)+r^{14}sin^{14} (\theta )}}{(r^{2}cos^{2}(\theta)+r^{2}sin^{2})^{3}(\theta)}=lim_{r\rightarrow0}\frac{4\sqrt{r^{14}}\sqrt{cos^{14}(\theta)+sin^{14}(\theta)}{}}{(r^{2}(cos^{2}(\theta)+sin^{2}(\theta))^{3}}=lim_{r\rightarrow0}\frac{4r^{7}\sqrt{cos^{14}(\theta)+sin^{14}(\theta)}}{r^{6}}=lim_{r\rightarrow0}4r\sqrt{cos^{14} (\theta)+sin^{14}(\theta)}[/itex]

    I don't see what I did wrong here.

    Also why don't I put [itex]\theta\rightarrow0[/itex]? I understand that [itex]\theta[/itex] goes in a period but shouldn't it approach anything? Maybe [itex]2\pi[/itex]?
     
  7. Jan 24, 2012 #6

    Dick

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    As along as r->0 the limit is still 0. The trig term bounded by [itex]\sqrt 2[/itex] for any value of [itex]\theta[/itex]
     
  8. Jan 24, 2012 #7
    What do you mean by that?

    Thank you for all the help. :smile:
     
  9. Jan 24, 2012 #8

    Dick

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    I mean that [itex]\sqrt(\sin^{14} \theta+\cos^{14} \theta)[/itex] is bounded as theta goes from 0 to 2*pi. It never gets larger than sqrt(2). If it went to infinity for some value of theta that could be a problem for the limit. The value of your function is between 0 and 4*r*sqrt(2). So as r->0 it must go to zero.
     
  10. Jan 24, 2012 #9
    I get it now. Thank you.
     
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