Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? Suppose a function $f$ is defined as follows: 
$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$
Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I am unable to find a counterexample. But I do not see how to prove continuity in general.
 A: Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$. 
Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$. 
Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$. 
From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4+y^4}\right| \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$ 
Now, multiply both sides by $|x|$ to get $\left|\dfrac{x^3y^2}{x^4+y^4}\right| \le \dfrac{1}{2}|x|$. 
Can you finish the problem from here?

 The Squeeze Theorem will be useful.

A: A fairly efficient way to approach this problem is to transform to polar coordinates and write
$$\lim_{(x,y)\to (0,0)}\frac{x^3y^2}{x^4+y^4}=\lim_{r\to \infty}\left(r\,\,\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right)$$
Noting that we can write 
$$\begin{align}
\left|\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right|&=\left|\frac{\cos^3(\phi)\sin^2(\phi)}{2(\sin^2(\phi)-\frac12)^2+\frac12}\right|\\\\
&\le 2
\end{align}$$
then the limit of interest is $0$.  Therefore, the function $f(x,y)$ is continuous at the origin.
A: Hint: use arithmetic-geometric mean inequality to show the function is continuous at $(0,0)$: for $a,b>0$, 
$$  ab \leq \frac{a^2 + b^2}{2} $$ 
