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Math Help - Proving Continuity

  1. #1
    Member Haven's Avatar
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    Proving Continuity

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

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

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

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

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

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

    \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 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 (x, y) \to (0,0).
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