Limits in polar coordinates I cannot understand why the following statement is false:
"Let $f$ be a function defined as $f:R^2\to R$ such that $f(0,0)=1$.  If for all $\varphi\in[0,2\pi[$ fixed we have $\lim_{r\to 0}f(rcos(\varphi),rsin(\varphi))=1$ then $f$ is continuous at $(0,0)$."
To me it seems that this is exactly the same concept as taking a limit in polar coordinates.  In this case we would normally see if the limit exists independently of $\varphi$.  Isn't this what we are doing here?
 A: By definition, $f$ is continuous at the point (0,0) if $f(x,y)\rightarrow f(0,0)$ as $(x,y)\rightarrow(0,0)$ along any $\textit{path}$.  The statement above says that $f(x,y)\rightarrow f(0,0)$ as $(x,y)\rightarrow(0,0)$ along any straight $\textit{line}$.  But there are other paths besides straight lines and the statement doesn't say anything about what happens to $f$ in these cases.  
A: The problem is that $f$ converges "at different rates" along lines passing through the origin. There is no lower bound to how slow $f$ converges on each line, so $f$ may not be continuous along a trajectory intersecting all rays originating from $(0,0)$.
Let $g:\mathbb{R} \to \mathbb{R}$ be the Conway base 13 function.
Define $f$:
$$f(r,\phi) = 1 + r^2 g(\phi)
$$
Evidently $\lim_{r \to 0} f(r,\phi) = 1$ for every $\phi$. (If you want a Cartesian represetation, replace $r$ with $\sqrt{x^2 + y^2}$, $\phi$ with $\operatorname{atan2}(y,x)$.)
but let $r(t) = e^{-t}, \phi(t) = t$. Then $\lim_{t \to \infty} (r(t) \cos \phi(t), r(t) \sin \phi(t)) = (0,0)$. However,
$$ \lim_{t \to \infty} f(r(t),\phi(t)) = \lim_{t \to \infty} 1 +e^{-2t}g(t)
$$
which does not converge.
