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Thread: Proving Continuity

  1. #1
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    Proving Continuity

    Use the fact $\displaystyle 4x^4y^2 \leq (x^4+y^2)^2$ to show that
    $\displaystyle
    f(x) = \left\{
    \begin{array}{cc}
    x^2+y^2-2x^2y - \frac{4x^6y^2}{(x^4+y^2)^2} & (x,y)\neq (0,0)\\
    0 & (x,y) = (0,0)
    \end{array}
    \right.
    $
    is continuous. I having troubles getting the algebra to work. Thanks in advance.
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  2. #2
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    $\displaystyle \displaystyle 4x^4y^2 \leq (x^4 + y^2)^2$

    $\displaystyle \displaystyle \frac{1}{(x^4 + y^2)^2} \leq \frac{1}{4x^4y^2}$, and since $\displaystyle \displaystyle 0 \leq \frac{1}{(x^4 + y^2)^2}$, that means

    $\displaystyle \displaystyle 0 \leq \frac{1}{(x^4 + y^2)^2} \leq \frac{1}{4x^4y^2}$

    $\displaystyle \displaystyle -\frac{1}{4x^4y^2} \leq -\frac{1}{(x^4 + y^2)^2} \leq 0$

    $\displaystyle \displaystyle -\frac{4x^6y^2}{4x^4y^2} \leq -\frac{4x^6y^2}{(x^4 + y^2)^2} \leq 0$

    $\displaystyle \displaystyle -x^2 \leq -\frac{4x^6y^2}{(x^4 + y^2)^2} \leq 0$

    $\displaystyle \displaystyle x^2 + y^2 - 2x^2y - x^2 \leq x^2 + y^2 - 2x^2y - \frac{4x^4y^2}{(x^4 + y^2)^2} \leq x^2 + y^2 - 2x^2y$

    $\displaystyle \displaystyle y^2 - 2x^2y \leq x^2 + y^2 - 2x^2y - \frac{4x^4y^2}{(x^4 + y^2)^2} \leq x^2 + y^2 - 2x^2y$.


    Now see what happens when you make $\displaystyle \displaystyle (x, y) \to (0,0)$.
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