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Math Help - Continuity and differentiability

  1. #1
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    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!
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  2. #2
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    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) .
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  3. #3
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    So it's not continuous in (0,0) but what about the other points that satisfies x=y^3? Thanks
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  4. #4
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    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.
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  5. #5
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    Thanks, so I can conclude that the function is not differentiable for x=y^3 because it's not continuous, right?
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  6. #6
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    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.
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  7. #7
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    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?
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  8. #8
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    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?
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  9. #9
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    That one is continuous because using polar coordinates results 0 no matter what, am I right?
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