1. ## Proving Continuity

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.

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

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