Showing continuity of $g(x, y)$ Let $g : \mathbb{R}^2 \to \mathbb{R}$. Let the domain be $y^2 \leq \frac{x}{3}$. 
$g(x, y) = \begin{cases}
\frac{x^3}{y^2} \left (\sqrt{x} - \sqrt{y^2 + x} \right ) & y \neq 0\\
0 & y = 0
\end{cases} $
I want to show $g$ is continuous at $(0, 0)$.
We have $|g(x, y) - g(0,0)| = \frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )$. 
Fix some $\epsilon > 0$. 
Now, I want to find a $\delta > 0$ such that $x^2 + y^2 < \delta$ $\implies $ $\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right ) < \epsilon$ 

$|\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )|=|\frac{x^3}{\sqrt{x}+\sqrt{y^2+x}}| \leq |x|^{\frac{5}{2}}$ 
So the limit of the function at zero is zero.
