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Math Help - Multivariable Delta Epsilon Proof

  1. #1
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    Multivariable Delta Epsilon Proof

    Q: limit of f(x,y)=(x^3-y^3)/(x^2+y^2) as (x,y) -> (0,)?

    What I tried:
    1. Picked 0 as L
    2. We know x^2+y^2 < d^2
    3. and lxl=<d and lyl=<d

    I can't think of an inequation needed

    Any help would be greatly appreciated. Thanks
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  2. #2
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    Quote Originally Posted by 6DOM View Post
    Q: limit of f(x,y)=(x^3-y^3)/(x^2+y^2) as (x,y) -> (0,)?

    What I tried:
    1. Picked 0 as L
    2. We know x^2+y^2 < d^2
    3. and lxl=<d and lyl=<d

    I can't think of an inequation needed

    Any help would be greatly appreciated. Thanks
    I think you mean inequality. As

     <br />
- \left(x^2+y^2\right)^{1/2} \le x,y \le \left(x^2+y^2\right)^{1/2}<br />

    try

     <br />
- \left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2}.<br />
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  3. #3
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    Quote Originally Posted by Danny View Post
    I think you mean inequality. As

     <br />
- \left(x^2+y^2\right)^{1/2} \le x,y \le \left(x^2+y^2\right)^{1/2}<br />

    try

     <br />
- \left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2}.<br />
    I'm not sure I understand you.

    I thought we needed some kind of inequality that looks like

    l(x^3-y^3)/(x^2+y^2)l =<

    so that we can set delta
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  4. #4
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    Quote Originally Posted by 6DOM View Post
    I'm not sure I understand you.

    I thought we needed some kind of inequality that looks like

    l(x^3-y^3)/(x^2+y^2)l =<

    so that we can set delta
    If

    <br />
\left(x^2+y^2\right)^{3/2} \le x^3 - y^3 \le \left(x^2+y^2\right)^{3/2},

    then

    <br />
\left(x^2+y^2\right)^{1/2} \le \frac{x^3 - y^3}{x^2+y^2} \le \left(x^2+y^2\right)^{1/2},

    then

    <br />
\left|\, \frac{x^3 - y^3}{x^2+y^2} \, \right| \le \left(x^2+y^2\right)^{1/2}.
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