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Continuity of multivariable functions question

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Is the function f(x,y) defined by
    f(x,y) = (yx^3 - 3y^3)/(x^2 + y^2), (x,y)!=(0,0)
    =0, (x,y)=(0,0)
    continuous everywhere in R^2? Give reasons for your answer.
    2. Relevant equations

    3. The attempt at a solution
    I changed f(x,y) into polar coordinates and found the limit as r->0, which is 0. I also found that f(x,0) = 0, f(0,y) = -3y^2 -> 0 as y->0, and f(x,x)= -infinity. However I'm not sure if I've done the question right or if this is enough working.
     
  2. jcsd
  3. May 30, 2010 #2
    Someone help!!!
     
  4. May 30, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi QuanticEnigma! Welcome to PF! :smile:

    (have a ≠ and a ∞ and a θ and try using the X2 tag just above the Reply box :wink:)
    Using polar coordinates is good, but it's not enough to check the limit only along the two axes.

    Hint: use the fact that cosθ and sinθ are always ≤ 1. :wink:
     
  5. May 30, 2010 #4
    Ok, I've decided to cheat and use Mathematica to plot the function, and I can see that the function is continuous everywhere :tongue:
    I think there must be some problem in my working, as the limits as f tends to (0,0) along the x and y axes is not the same as the limit along the line y=x...I'm confused
     
  6. May 30, 2010 #5

    tiny-tim

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    uhh? :confused: isn't f(x,x) = (x2 - 3x)/2 ?

    anyway, try what I said about θ. :wink:
     
  7. May 30, 2010 #6
    Yeah it is, my bad (hope I don't make that mistake in the exam :tongue:)

    I don't really know what you mean by theta, I've already worked out the limit of f (in polar coordinates) as r tends to 0 to be zero, and I thought this implied that any line approaching (0,0) would have limit zero...I'm not sure though :confused:

    Oh and by the way, every time I input latex code it always outputs f(x,y), even when I've copied and pasted it from other threads, any idea on how to fix this?
     
  8. May 30, 2010 #7

    tiny-tim

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    I'm dubious about what you mean by "as r tends to zero".

    I haven't seen your exact proof, but I assume you have an f(r,θ), and you can't prove continuity by only letting r -> 0 any more than you could by only letting x -> 0 and y -> 0 (separately).
    It's a problem which has affected Preview ever since the last server update … you can fix it by clicking "Refresh" or "Reload" on your browser menu (do a tag search for "preview" for details).

    Or just use the X2 tag. :wink:
     
  9. May 30, 2010 #8
    Ok, I have
    [tex]
    f(x,y)=\frac{r^3\cos^3(\theta)r\sin(\theta)-3r^3\sin^3(\theta)}{r^2\cos^2(\theta)+r^2\sin^2(\theta)} = \frac{r^3(r\cos^3(\theta)\sin(\theta)-3\sin^3(\theta)}{r^2}=r(r\cos^3(\theta)\sin(\theta)-3\sin^3(\theta))
    [/tex]

    Since I have an r out the front of everything, i didn't worry about theta, because when r=0, f(r,[itex]\theta[/itex]) would equal zero.
     
  10. May 30, 2010 #9

    tiny-tim

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    (what happened to that θ i gave you? :confused:)

    But that doesn't work if the inside of the bracket can be "infinitely" large, so you still have to prove that it can't be. :wink:
     
  11. May 30, 2010 #10
    Sorry, it's supposed to be [tex]f(r,\theta)[/tex] not [tex]f(x,y)[/tex]
     
  12. May 30, 2010 #11
    Would sin[tex]\theta[/tex] and cos[tex]\theta[/tex] being [tex]\leq[/tex] 1 mean that the expression inside the bracket is between -1 and 1?
     
  13. May 30, 2010 #12

    tiny-tim

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    Yup! :biggrin:

    (well, between ±(r±3) :wink:)
     
  14. May 30, 2010 #13
    Sweet, thanks for the help :smile:
     
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