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Math Help - Show the the derivative exists but not continuous

  1. #1
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    Show the the derivative exists but not continuous

    Hi dear friends here, I got a question, it supposed to be quite simple, but it is a little bit tricky, can anyone please help me? Thanks a lot.
    Here it is:


    f(x,y)={x^3/(x^6+y^2), for (x,y) does not equal (0,0)}
    f(x,y)={0, for (x,y)=(o,o)}
    Show that df/dx exists at (0,0) but is not continuous there.


    I used first principle to prove the limit of df/dx is 0 at (x,y)=(0,0), which means it does exist, am I right? What is the exact definition to say something exists?
    Then, I can imagine it is discontinuous because at (0,0) the graph of df/dx should be broken, which has the circle there, but I just forgot how to show it. I thought the definition of "continuiy" is the value of limit equals to the value of function. So how do I write about it? Please help me, thanks a lot.
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  2. #2
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    Quote Originally Posted by tsang View Post
    Hi dear friends here, I got a question, it supposed to be quite simple, but it is a little bit tricky, can anyone please help me? Thanks a lot.
    Here it is:


    f(x,y)={x^3/(x^6+y^2), for (x,y) does not equal (0,0)}
    f(x,y)={0, for (x,y)=(o,o)}
    Show that df/dx exists at (0,0) but is not continuous there.


    I used first principle to prove the limit of df/dx is 0 at (x,y)=(0,0), which means it does exist, am I right? What is the exact definition to say something exists?
    You mean \frac{\partial f}{\partial y} here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

    Yes, thi partial derivative exists at x= 0 if and only if \lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h} exists.

    In this case, that limit is \lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7} = \lim_{h\to 0}\frac{1}{h^4}. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?

    Then, I can imagine it is discontinuous because at (0,0) the graph of df/dx should be broken, which has the circle there, but I just forgot how to show it. I thought the definition of "continuiy" is the value of limit equals to the value of function. So how do I write about it? Please help me, thanks a lot.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You mean \frac{\partial f}{\partial y} here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

    Yes, thi partial derivative exists at x= 0 if and only if \lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h} exists.

    In this case, that limit is \lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7} = \lim_{h\to 0}\frac{1}{h^4}. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?


    Thank you. I'm sure I copied down the right question. Also, I do mean by \frac{\partial f}{\partial y}, sorry that I'm not very good at with using code.
    I think the question is trying to say \frac{\partial f}{\partial y} is exists but discontinues, not original function. So what I did was differentiate original function first, and I end up with [tex]\frac{\partial f}{\partial y}=(3x^2y^3-3x^8y)/((x^6+y^2)^2)[tex], and use first principle as you did, so I can find the limit equal to 0.
    But not sure how to sure discontinues.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    You mean \frac{\partial f}{\partial y} here, don't you? That's very different from saying "the derivative exists at (0, 0)" or "the function is differentiable at (0, 0)"

    Yes, thi partial derivative exists at x= 0 if and only if \lim_{(h,k)\to 0,0}\frac{f(0+h, 0)- f(0, 0)}{h} exists.

    In this case, that limit is \lim_{h\to 0}\frac\frac{{(0+h)^3}{(0+h)^6+ 0^2}- 0}{h}= \lim_{h\to 0}\frac{h^3}{h^7} = \lim_{h\to 0}\frac{1}{h^4}. It does NOT look to me like that limit exists. Are you sure you have copied the problem correctly?

    thank you, I'm sure I got right question, and it is \frac{\partial f}{\partial y}, sorry that I'm not very good with using code.
    I think the question is asking about \frac{\partial f}{\partial y}, rather than original function. I differentiate original function first, and I end up with df/dx=(3x^2y^3-3x^8y)/(x^6+y^2)^2, which can find limit at (0,0) is 0.
    But I'm not sure how to show it is discontinues.
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