<s>[INST] Proof coutinuity $f:=\left\{\begin{matrix}
\frac{xy}{\sqrt{x}+y^2}, x,y\neq 0\ begin & \\ 
0, x,y=0 & 
\end{matrix}\right.$
Is f continuous in (0,0)?
My idea is:
$\left | f(x,y) \right | = \left | \frac{x*y}{\sqrt{x}+y^2} \right |\leq \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right|$
we know:
$\sqrt{x}+\sqrt{y}>\sqrt{x+y} \implies \left | \frac{x*y}{\sqrt{x}+\sqrt{y}} \right| \leq\left | \frac{x*y}{\sqrt{x+y}} \right|$
we know:
$\sqrt{x+y} \geq 2*\sqrt{x}\sqrt{y} \implies \left | \frac{x*y}{\sqrt{x+y}} \right| \leq \left | \frac{x*y}{(x*y)^{\frac{1}{4}}} \right|=(x*y)^{\frac{3}{4}}$
with $(x,y) \implies 0$ we get $(x*y)^{\frac{3}{4}}$ =0
Can someone please check my suggestion?
 [/INST] We assume that $x\ge 0$.  Now, note that 
$$\sqrt{x}+y^2\ge\sqrt{x} \implies \frac{1}{\sqrt{x}+y^2}\le \frac{1}{\sqrt{x}}\implies \frac{xy}{\sqrt{x}+y^2}\le \sqrt{x}y \to 0$$
</s>