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Math Help - Limit and differentiability

  1. #1
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    Limit and differentiability

    Let

    f(x, y) = \sqrt[3]{xy^2 + x^2 y}.

    By using the definition of differentiability show that f is not differentiable at (0,0).

    Definition is:

     <br />
\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\Delta f - f_x(x_0,y_0)\Delta x -f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}=0<br />

    I already know that f_x (0,0)=f_y(0,0)=0. Also the increment is \Delta f = (0+\Delta x, 0+ \Delta y) = \sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}. So

    \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}

    Now, I need to evaluate this to show that it doesn't =0. I'm stuck here. I don't know if I should use the conjugate OR use polar coordinates to evaluate this limit? Can someone help?
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  2. #2
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    Quote Originally Posted by demode View Post
    Let

    f(x, y) = \sqrt[3]{xy^2 + x^2 y}.

    By using the definition of differentiability show that f is not differentiable at (0,0).

    Definition is:

     <br />
\lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\Delta f - f_x(x_0,y_0)\Delta x -f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}=0<br />

    I already know that f_x (0,0)=f_y(0,0)=0. Also the increment is \Delta f = (0+\Delta x, 0+ \Delta y) = \sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}. So

    \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}

    Now, I need to evaluate this to show that it doesn't =0. I'm stuck here. I don't know if I should use the conjugate OR use polar coordinates to evaluate this limit? Can someone help?

    Put \Delta_y=m\Delta_x and you'll get the limit depends on m... and correct the expression for f(\Delta_x,\Delta_y) , it should be \sqrt[3]{\Delta_x\Delta^2_y+\Delta^2_x\Delta_y} .

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Put \Delta_y=m\Delta_x and you'll get the limit depends on m... and correct the expression for f(\Delta_x,\Delta_y) , it should be \sqrt[3]{\Delta_x\Delta^2_y+\Delta^2_x\Delta_y} .

    Tonio
    So

    \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(\Delta y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}

    becomes

    \lim_{(\Delta x,m \Delta x) \to (0,0)} \frac{\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}}{\sqrt{(\Delta x)^2+(m \Delta x)^2}}

    Again I don't know how to evaluate this!
    Last edited by demode; March 13th 2010 at 09:19 PM.
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  4. #4
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    Would it not have been easier to convert to polar coordinates, so that the limit will be in terms of "r"?
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  5. #5
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    Quote Originally Posted by demode View Post
    So

    \lim_{(\Delta x,\Delta y) \to (0,0)} \frac{\sqrt[3]{\Delta x(\Delta y)^2 + (\Delta x)^2 \Delta y}}{\sqrt{(\Delta x)^2+(\Delta y)^2}}

    becomes

    \lim_{(\Delta x,m \Delta x) \to (0,0)} \frac{\sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}}{\sqrt{(\Delta x)^2+(m \Delta x)^2}}

    Again I don't know how to evaluate this!

    You need some high school algebra here to simplify that stuff...for example, \sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}=\Delta_x\sqrt[3]{m^2+m} , etc.

    Tonio
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    Quote Originally Posted by tonio View Post
    You need some high school algebra here to simplify that stuff...for example, \sqrt[3]{\Delta x(m \Delta x)^2 + (\Delta x)^2 m \Delta x}=\Delta_x\sqrt[3]{m^2+m} , etc.

    Tonio
    yes

    \lim_{m \to (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}

    Since this is a limit involving radicals, I need to rationalize the denominator:

    \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}

    This is where I'm stuck. I know the denominator will be 1+m, but how do we simplify the numerator?
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    Quote Originally Posted by demode View Post
    yes

    \lim_{m \to (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}


    Why do you take the limit when m --> (0,0) ?? This doesn't even make sense! The limit is when (\Delta_x,\Delta_y)\rightarrow (0,0) ...!

    Tonio

    Since this is a limit involving radicals, I need to rationalize the denominator:

    \lim_{m \to (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}

    This is where I'm stuck. I know the denominator will be 1+m, but how do we simplify the numerator?
    .
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  8. #8
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    Yes but then how are we going to evaluate this limit?

    I mean if

     <br />
\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}<br />

    And in the expression on the RHS there are no more \Delta_x, \Delta_y terms left, and we don't know the value of m! So, what do I need to do?

    Of course, I could rationalize the denominator but I don't know how to simplify the numerator in the following:

     <br />
\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}<br />
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  9. #9
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    I even tried writing out the powers as a fraction, that is, to let the square roots be ^(1/2) and such. But still I end up with m's and you didn't mention what the value of m was! How am I supposed to show that the limit doesn't equal zero?
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  10. #10
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    You actually do know the value of m - you can set it to be anything you like!
    That is why the function is not differentiable around (0,0) - because the limit depends on m (ie. you can get two different values for the limit for two different values of m) and as a result, it does not exist (do you understand why?)
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  11. #11
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    Quote Originally Posted by demode View Post
    Yes but then how are we going to evaluate this limit?

    I mean if

     <br />
\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\Delta_x\sqrt[3]{m^2+m}}{\Delta x \sqrt{1+m^2}}= \lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}}<br />

    And in the expression on the RHS there are no more \Delta_x, \Delta_y terms left, and we don't know the value of m! So, what do I need to do?


    You need to really concentrate a little more: if you put \Delta_y=m\Delta_x , then (\Delta_x,\Delta_y)\rightarrow (0,0) \Longleftrightarrow \Delta_x\rightarrow 0 , so THE LIMIT IS \frac{\sqrt[3]{m^2+m}}{\sqrt{1+m^2}} , and then

    the limit depends on m (unless you could prove the limit is the same no matter what m we choose...which of course you won't be able to prove since it isn't true ).

    Tonio



    Of course, I could rationalize the denominator but I don't know how to simplify the numerator in the following:

     <br />
\lim_{(\Delta_x,\Delta_y)\rightarrow (0,0)} \frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}<br />
    .
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