question about continuity: using polar coordinates

Question

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.

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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 $\cos^2\theta+\sin^2\theta=1$.

Answer 1

The use of polar coordinates and the limit $r\rightarrow 0$ to compute $\lim_{(x,y)\rightarrow (0,0)}f(x,y)$ is not a choice of path to approach your limit point.
With polar coordinates you are just "parametrizing" all the point in the neighborhood of $(0,0)$ through their distance from $(0,0)$ and their angle w.r.t. the positive $x$-axis. The limit $r\rightarrow 0$ is along any path connecting the starting point and $(0,0)$, not (only) the straight lines. If such limit depends on the path itself, like in the case of the limit

$$\lim_{(x,y)\rightarrow (0,0)}\frac{xy}{x^2+y^2}$$

then we say that the function is not continuous at the limit point.

Of course, the use of polar coordinates is great for some functions, while it can be a bad choice for many others. To directly find a counterexample to continuity (if it exists!) is probably the best way to answer to such kind of problems.

Answer 2

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' $(0,0)$.