Question about continuity in polar coordinate

Question

Let $x=r\cos(\theta)$ et $y=r\sin(\theta)$ and $f(0,0)=0$. Suppose I've shown that $$ \left|f\left(x,y\right)-f\left(0,0\right) \right| \leq \left|\sin^3\left(\theta\right)\right| $$ Can I conclude that $f$ is continuous ? Meaning, does I have $$\sin^3\left(\theta\right)\underset{(x,y) \rightarrow 0}{\rightarrow}0 \ ?$$

Answer 1

Actually, you cannot conclude that $f(x,y) \to 0$, Since the limit of $f(x,y)$ relied on the $\theta$. Here is a counterexample:

$$ f(x,y)= \frac{xy}{x^2+y^2}.$$

$\lim_{(x,y)\to 0} f(x,y) = \cos \theta \sin \theta $, the limit is rely on the $\theta$ even though $r$ is vanish.

Answer 2

You can have $(x,y)$ tend to $(0,0)$ just by making $r$ smaller and smaller without acting on $\theta$. So the answer is no - more precisely, $\sin^3 \theta$ does not have a limit at all as $(x,y)\to (0,0)$.