1. ## 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.

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

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