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Limit of a 3d function

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data

    f(x,y) is a hybrid function

    1. ( ( xy+x )^3 ) / ( x^2 + xy ) when (x,y) not equal to (0,0)
    2. 0......................................when (x,y) = (0,0)

    Is this function continuous at the origin?

    2. Relevant equations

    .......lim.......df |(x,y) = df(0,0)
    (x,y)->(0,0)

    3. The attempt at a solution

    To determine this I have to use the continuity Rule (above) and first principle.

    so

    lim....f(0,h) - F(0)........(Imagine the divisor line there)
    h->0.........h..............(and excuse the dots, spaces don't work)

    =
    lim.....[x(y+x^2) - 0 ] x [ 1 ]
    h->0..[...x(x+y).......]....[ h ]
    ................^

    From here I'm a little stuck.
    Do I cancel the x and solve the limit which result in infinity and therefore the function is not continuous at the origin OR put in the value of (0,h) for (x,y) and get the limit to equal zero?

    ie.
    ....lim.....[0(h+0) - 0 ] x [ 1 ]
    ....h->0..[...0(0+h).....]....[ h ]

    ..=lim...........1
    ....h->0 .......h
     
  2. jcsd
  3. Mar 29, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm extremely confused. You say the problem is to determine whether or not the function is continuous at the origin and refer to the "continuity rule (above)" but I see no "continuity rule". what you have written:
    .......lim.......df |(x,y) = df(0,0)
    (x,y)->(0,0)
    has little to do with continuity.

    You then seem to be using the definition of derivative to try to find the partial derivative with respect to y at (0,0)- I see no reason to want to do that and, if you really want to find it, I don't see why you have not set (x, y) equal to (0, 0) in the calculations from the start.

    In any case, a function, f(x,y), is continuous at (0, 0) if and only if its limit, as (x, y) goes to (0, 0) is f(0,0).
    I would recommend changing to polar coordinates: [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex] so [itex](xy+ x)^3= (r^2cos(\theta)sin(theta)+ r cos(theta))^3= r^3cos^3(\theta)(r sin(\theta)+ 1)^3[/itex] and [itex]x^2+ xy= r^2cos^2(\theta)+ r^2sin(\theta)cos(\theta)= r^2cos(\theta)(cos(\theta)+ sin(\theta))[/itex]. Since (x, y) is not (0, 0), r is not 0 so we can cancel the "r"s. The fraction reduces to
    [tex]r\frac{cos^3(\theta)(rsin(\theta)+ 1)^3}{cos(\theta)(cos(\theta)+ sin(\theta))}[/tex]
    That goes to 0 as r goes to 0, regardless of what [itex]\theta[/itex] is so the function is continuous at (0,0).

    If your question was really whether or not the partial differentials are continuous at the origin, that's a different question.
     
  4. Mar 29, 2008 #3
    Yea, I think I got a little confused since my book is a mess.
    I had the question and answers messed up.
    The continuity Rule should of been f(x) not df and in your exact words
    "f(x,y), is continuous at (0, 0) if and only if its limit, as (x, y) goes to (0, 0) is f(0,0)."
    I took that from another page in my book to see if the partial derivative was continuous. :S

    1.To calculate the partial derivative at (0,0)
    2.To show f(x,y) is continuous

    1. I think I have to go to first principle to calculate the partial derivative.
    2. Use the squeeze theorem to show the limit is zero then it's continuous.
    ( haven't done much on polar yet but I understand what you have done)

    Thanks for your help.
     
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