Results 1 to 9 of 9

Math Help - Continuity and differentiability

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    13

    Continuity and differentiability

    Hi all,

    I got some problems doing some exams exercises...
    They ask me to study the continuity and the differentiability of some functions but I always get similar results...

    f(x,y)=x/(x-y^3), if x !=  y^3
    f(x,y)=0, if x=y^3

    The function is continuous when x !=  y^3 but it is not when x=y^3 because the limit for (x,y)->(y^3,y) results +∞ or -∞ depending on the approaching direction... Am I wrong? If yes, how do I have to solve that limit? Thanks to all!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Quote Originally Posted by jollysa87 View Post
    Hi all,

    I got some problems doing some exams exercises...
    They ask me to study the continuity and the differentiability of some functions but I always get similar results...

    f(x,y)=x/(x-y^3), if x !=  y^3
    f(x,y)=0, if x=y^3

    The function is continuous when x !=  y^3 but it is not when x=y^3 because the limit for (x,y)->(y^3,y) results +∞ or -∞ depending on the approaching direction... Am I wrong? If yes, how do I have to solve that limit? Thanks to all!
    Actually, f is not continuous. If we follow the path x = m y^3, m \ne 1 then

     <br />
\lim_{(x,y)->(0,0)} \frac{m y^3}{m y^3-y^3} = \frac{m}{m-1}<br />
which clear changes as we very  m. Since we get different limits f is not continuous at (0,0) .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2009
    Posts
    13
    So it's not continuous in (0,0) but what about the other points that satisfies x=y^3? Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Quote Originally Posted by jollysa87 View Post
    So it's not continuous in (0,0) but what about the other points that satisfies x=y^3? Thanks
    Still not. You have f defined as 0 if x = y^3. Thus, in a neighborhood of this curve, if continuous, then f can be made arbitrary close to 0. However, along x = my^3, the function is \frac{m}{1-m} which is large near m = 1, not zero.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2009
    Posts
    13
    Thanks, so I can conclude that the function is not differentiable for x=y^3 because it's not continuous, right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Quote Originally Posted by jollysa87 View Post
    Thanks, so I can conclude that the function is not differentiable for x=y^3 because it's not continuous, right?
    Yep.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2009
    Posts
    13
    So the following function too is not continuous for y=x?


    f(x,y)=y(1+x)/(x-y), if y!=x
    f(x,y)=0, if y=x

    Because if I follow the path y=mx, m!=1 then the limit change with m?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,324
    Thanks
    8
    Quote Originally Posted by jollysa87 View Post
    So the following function too is not continuous for y=x?


    f(x,y)=y(1+x)/(x-y), if y!=x
    f(x,y)=0, if y=x

    Because if I follow the path y=mx, m!=1 then the limit change with m?
    Yep - you got it!

    Now consider the following

     <br />
\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2+y^2}<br />

    Every path you follow you get 0. So, now what?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Jul 2009
    Posts
    13
    That one is continuous because using polar coordinates results 0 no matter what, am I right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Continuity and Differentiability
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 10th 2010, 01:32 AM
  2. differentiability and continuity
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: November 23rd 2010, 06:18 AM
  3. Differentiability and continuity
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 30th 2010, 03:26 AM
  4. continuity and differentiability
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 2nd 2009, 12:03 AM
  5. Continuity/Differentiability
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 2nd 2007, 03:33 PM

Search Tags


/mathhelpforum @mathhelpforum