Proving Continuity

**Haven** · March 20th 2011, 10:20 PM

Use the fact $4x^4y^2 \leq (x^4+y^2)^2$ to show that
$ 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.

**Prove It** · March 20th 2011, 10:49 PM

$\displaystyle 4x^4y^2 \leq (x^4 + y^2)^2$

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