Continuity of $\begin{cases}(xy+y^2)/(x^4+y^2)&\text{if }(x,y)\neq(0,0),\\0&\text{if }(x,y)=(0,0)\end{cases}$ at origin using polar coordinates Study the continuity of $$f(x,y)=\begin{cases}\dfrac{xy+y^2}{x^4+y^2}&\text{if }(x,y)\neq(0,0),\\0&\text{if }(x,y)=(0,0),\end{cases}$$ at $(x,y)=(0,0)$ using polar coordinates.

I know that $f(0,0)=0$ so, if $\lim_{(x,y)\to(0,0)}{f(x,y)}$ exists then it must be equal to $0$.
I want to prove that is not continuous at origin using polar coordinates. Let $(x,y)=(r\cos\theta,r\sin\theta)$. Then
$$
\lim_{(x,y)\to(0,0)}{f(x,y)}=\lim_{r\to0}{\frac{r^2\cos\theta\sin\theta+r^2\sin^2\theta}{r^4\cos^4\theta+r^2\sin^2\theta}}=\lim_{r\to0}{\frac{\cos\theta\sin\theta+\sin^2\theta}{r^2\cos^4\theta+\sin^2\theta}}=\frac{\cos\theta\sin\theta+\sin^2\theta}{\sin^2\theta}=1+\cot\theta,
$$
so, because the limit depends on the value of $\theta$, then the limit does not exist, hence $f(x,y)$ is not continuous at $(0,0)$.
Is that correct? Can we use polar coordinates here?
Thanks!

Although your argument contains a grain of truth, it is not quite correct as it is written, since you wrote that the limit $\lim_{(x,y)\to (0,0)} f(x,y)$, which doesn't exist, is equal to the limit $\lim_{r \to 0} (\cdots)$, which does exist (for $\sin \theta \neq 0$) if you just view it as an ordinary single-variable limit which depends parametrically on a constant $\theta$; in fact, you just computed it yourself like that and wrote that it's equal to $1 + \cot \theta$.
The problem is that you want $\theta$ to be able to vary independently of $r$ as $r \to 0$, so you can't treat $\theta$ as a constant here. It should be viewed as an arbitrary function $\theta(r)$.
In fact, polar coordinates are mainly useful for proving that a limit exist, namely if you can write $f(x,y)$ as a bounded factor times another factor which depends only on $r$ (no $\theta$!) and which tends to zero as $r \to 0$ (really just a single-variable limit here!), then $f(x,y)\to 0$ as $(x,y)\to(0,0)$.
To show that a limit does not exist, you instead find two ways of approaching the point such that you get two different values. In your case, consider $f(t,0)$ and $f(0,t)$ as $t\to 0$, for example.
So actually I don't quite know how I would like to write the argument in a nice way if someone forced me to use polar coordinates in order to show that a limit does not exist! I would probably just write $f(x,y)$ in polar coordinates first,
$$
f(r \cos\theta(r), r \sin\theta(r)) = \cdots,
$$
(no “$\lim$” here) and then say that by making different choices of the function $\theta(r)$ (for example different constant functions!), you can make $f$ approach different values. And I would give examples of two such function $\theta(r)$ which give different limits for $f(r \cos\theta(r), r \sin\theta(r))$ as $r \to 0$.
