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Homework Help: Limits of Multivariable Functions

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the limit, if it exists, or show that the limit does not exist.

    limit (x,y) --> (0,0)

    a) f(x,y) = (xycosy) / (3x^2 + y^2)

    b) f(x,y) = (xy) / sqrt(x^2 + y^2)

    c) f(x,y) = ((x^2)ye^y) / ((x^4) + 4y^2)

    2. Relevant equations

    3. The attempt at a solution

    a) For this one, I did (0,y)-->(0,0) and got 0; then did (x,0) --> (0,0) and got 0. Then I substituted y=x, so (x,x) --> (0,0). I ended up getting ((x^2)cosx) / 4x^2; which I evaluated as x goes to zero, the limit would not exist (since the bottom would be zero).

    b) Again I did x=0, and y=0 and came up with 0 for both of those. Would I then, substituted y=x again? Ending up with x^2 / xsqrt(2), so x/sqrt(2) goes to zero as x goes to zero. So the limit would be zero. Would I have to try more paths?

    c) Again, x=0, y=0 resulted in limit of 0. Next I attempted x = sqrt(y); so ((y^2)e^y) / ((y^2) + 4y^2) which lead to (1/5)e^y. So as y goes to zero, e^y goes to 1, and the limit goes to 1/5. Therefore, the limit does not exist.

    Did I go about these in the right way, and would I need to test more paths for each of them?
  2. jcsd
  3. Oct 28, 2007 #2


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    Homework Helper
    Gold Member

    You need to show it for all paths even y=x^2 or even y=sin(x).
  4. Oct 29, 2007 #3
    For all three problems?
  5. Oct 29, 2007 #4


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    Science Advisor

    Showing that you get the same limit for some paths is not enough- there still might be some, perhaps very complex, path for which you would get a different result. JasonRox is correct that you would have to show that you get the same limit for every possible path but, in general, there is no good way to do that.

    I would recommend changing into polar coordinates. That way the distance from (0,0) is measured only by the variable r. If the limit, as r goes to 0, does not depend on [itex]\theta[/itex] then that is the limit of the whole function. If that limit does depend on [itex]\theta[/itex] then the limit does not exist. Leave things like cos(y) and ey in terms of y and hope you can decide without changing that!
  6. Oct 29, 2007 #5
    I've never really been any good at converting from Cartesian to Polar (for some reason my college just doesn't seem to do it that often). So I'm not really sure if I'm doing this right.

    a) converted the first equation into (r, [tex]\theta[/tex]) --> (0,0)

    (rcos[tex]\theta[/tex]rsin[tex]\theta[/tex]cosy) / ((r^2)(3(cos([tex]\theta[/tex])^2) + sin([tex]\theta[/tex])^2))

    which gave me

    (cos[tex]\theta[/tex]sin[tex]\theta[/tex]cosy) / (3(cos([tex]\theta[/tex])^2) + sin([tex]\theta[/tex])^2)

    From here, where do I go? Or would I just say that since the r's cancelled it's only dependent upon [tex]\theta[/tex], which means it does not exist?

    b) (rcos[tex]\theta[/tex]rsin[tex]\theta[/tex]) / r = cos[tex]\theta[/tex]sin[tex]\theta[/tex] limits goes to zero?

    c) This one I have no clue where to go, here's what I attempted, I converted it to (r, [tex]\theta[/tex]):

    ((r^2)(cos([tex]\theta[/tex])^2)rsin[tex]\theta[/tex](e^y)) / ((r^4)(cos([tex]\theta[/tex])^4) + 4(r^2)sin([tex]\theta[/tex])^2)

    then I reduced it to

    (r(cos([tex]\theta[/tex])^2)sin([tex]\theta[/tex])(e^y)) / ((r^2)(cos([tex]\theta[/tex])^4) + 4(sin([tex]\theta[/tex])^2))

    Then I get stuck, what do I do from there? =/
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