Results 1 to 8 of 8

Math Help - Functions of two variables continuity question

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    2

    Functions of two variables continuity question

    Consider the following functions each of which is defined on the x - y plane

    f1(x) = (x-y)/(x+y) if x + y is not 0 and otherwise f1(x,y) = 0

    f2(x,y) = (xy)/(x^2 + y^2) if (x,y) is not (0,0) and otherwise f2(0,0) = 0

    f3(x,y) = (x^3 - y^3)/(x^2 + y^2) if (x,y) is not (0,0), and otherwise f3(0,0) is 0

    Which of these is continuous

    A) none B) f1 only C) f2 only D) f3 only E) all three

    I know the defination of continuity for a single variable is the lim as x-> a of f(x) = f(a)

    So i assume for two variables it should be the lim as (x,y) -> (a,b) of f(x,y) = f(a,b)

    But I am not sure how to figure this out can someone help out please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Its pretty easy, in each function you have f(0,0)=0

    Take the limit for each one as (x,y) \to (0,0)

    If the limit = 0 , then the function is continuous at (0,0)

    Do you have any problems in evaluating these limits?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    4
    But the lim (x,y) ->(0,0) is 0 for all them is it.. I mean Since the function is peicewise defined... If i plug in (0,0) then each one gives 0/0 Thats what i do not understand.

    I understand it is in indeterminant form. But we haven't used polar coordinates. As far as algebra goes, nothing cancels when factored.

    With the path rule should I approach (0,0) from the left and (0,0) from the right? I am not sure what that will do..
    Last edited by ldaniels241; April 15th 2010 at 08:53 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    You should know that 0/0 is undefined
    the limit still exist or does not exist
    we do not know anything about it
    you should use another methods to evaluate it
    such as: polar coordinates, 2-path rule , algebric methods .. etc
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    2
    Well since it is in indeterminant form in there a l'hopitals rule for a function of two variables? Because I am not sure how to do a path argument
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    No.
    There is no L'Hospital's Rule for the multivariable limits.

    For the first one, when y=0 , the limit = 1. But when y=x, the limit=0.
    So the limit does not exist.
    Hence, f1 is not continuous at (0,0) ..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2010
    Posts
    4
    I see I undersand. You found two paths in which the limits are not equal so the limit does not exist because if the limit did exist, we would get the same limit no matter what path we took.

    So for f2(x,y) = xy/(x^2 + y^2)

    The path y = x gives me the limit = 1/2

    But the path y = 0 gives me the limit = 0

    This means f2(x,y) is not continuous at (0,0) either

    And for f3(x,y) = x^3 - y^3/(x^2 + y^2)

    the path y = x gives me the limit = 0
    the path y = 1 gives me the limit = -1

    So f3(x,y) is not continuous at (0,0) either. So none of them are.

    is this correct?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Correct.
    But you should know that this method does not work if the limit exist.

    As an example: \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2+y^2}}

    If you take any path, the limit always will be 0
    Here, you should think about another way to deal with this limit
    What I want to say is that do not think in every problem that the limit does not exist and try to search for paths that have different values !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: September 12th 2011, 06:26 PM
  2. Replies: 1
    Last Post: June 5th 2011, 04:57 PM
  3. Continuity of a function of 2 variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2010, 03:11 PM
  4. question on continuity of functions
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 12th 2009, 07:00 PM
  5. Continuity in 2 variables.
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 23rd 2009, 12:12 AM

Search Tags


/mathhelpforum @mathhelpforum