question about continuity: using polar coordinates Given a function $f\colon\mathbb R^2\rightarrow \mathbb R$ I want to study continuity. So I know the $\varepsilon-\delta$ and sequence criterion.
Now we had polar coordinates in lectures: set $x=r\cos\theta$ and $y=r\sin\theta$ and then consider $r\rightarrow 0$ for continuity in $(0,0)$.
This transformation seems very usefull for expressions like e.g. $\frac{xy^2}{x^2+y^2}$ but don't I approach the function only on all straight lines in $(0,0)^t$ ? And for continuity I have to approach $(0,0)^t$ however I want to which I don't do using polar coordinates. So why can I still use polar coordinates? Thanks for helping.

Add: I don't want a solution for the continuity of the example above. Clearly $|\frac{xy^2}{x^2+y^2}|\leq |y|<\epsilon$ and so continuity in $(0,0)$ with $f(0,0)=0$ which I also get with polar coordinates since $\cos^2\theta+\sin^2\theta=1$.

If you fix $\theta$ and just let $r\to 0^+$ then you are approaching $(0,0)$ only on straight lines. This can indeed be useful in order to show that a limit does not exist, i.e. providing two different values for $\theta$ which result in two different limits. If you want to cover every path that approaches $(0,0)$ and still use polar coordinates, then you need to consider $\theta$ as a function (e.g. $\theta=\theta(r)$ arbitrary), rather than a constant. In your example,
$$
\lim_{r\to 0^+}\frac{\big(r\cos\theta(r)\big)\cdot\big(r \sin\theta(r)\big)^2}{\big(r\cos\theta(r)\big)^2 + \big(r\sin\theta(r)\big)^2}
=
\lim_{r\to 0^+}\frac{r^3\cos\theta(r)\sin^2\theta(r)}{r^2}
=
\lim_{r\to 0^+}r\cos\theta(r)\sin^2\theta(r)
=
0
$$
Note that considering $\theta=\theta(r)$ rather than $r=r(s)$ and $\theta=\theta(s)$ with $r(s)\to0^+$ for $s\to 0^+$ is assuming you are somehow 'strictly approaching' $(0,0)$.
