Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
 A: Hint. By considering polar coordinates,
$$
x:=r\cos\theta,\quad y:=r\sin\theta,
$$ you get
$$
\begin{align}
f(r\cos\theta,r\sin\theta)&=\frac{r^5(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)}{r^4}\\\\
&=r(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)
\end{align}
$$ then observe that
$$
\left|\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta\right|\leq7
$$ gives

$$
\left|f(r\cos\theta,r\sin\theta)\right|\leq7r
$$ 

for all $r>0$ and $\theta$.
A: As an another approach:
$$\bigg|\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{x^4-4x^2y^2-y^4}{(x^2+y^2)^2}\bigg|=|x|\bigg|\frac{(x^2+y^2)^2-6x^2y^2-2y^4}{(x^2+y^2)^2}\bigg| \\ \le |x|\bigg(1+\bigg|\frac{6x^2y^2+2y^4}{(x^2+y^2)^2}\bigg|\bigg)\le |x|\bigg(1+\bigg|\frac{6x^2y^2+6y^4}{(x^2+y^2)^2}\bigg|\bigg) \\ =|x|\bigg(1+\bigg|\frac{6y^2}{x^2+y^2}\bigg|\bigg)\le |x|(1+6)=7|x|.$$
Therefore, the limit is zero as $x$ approaches zero.
A: It is easy to estimate for instance:
$$
\begin{aligned}
|x^5-4x^3y^2-xy^4|
&\le |x^5|+|-4x^3y^2|+|-xy^4| \\
&= |x|\cdot( x^4+4x^2y^2+y^4) \\
&\le |x|\cdot( \ (x^2+y^2)^2 + 2(x^2+y^2)^2 + (x^2+y^2)^2\ )\\
&= 4|x|\; (x^2+y^2)^2\ .
\end{aligned}
$$
So is $f$ denotes the given function we have $|f(x,y)|\le 4|x|$, which holds also in zero (which is $(0,0)$). This estimation leads to continuity in zero.
