direction limits and double limit Let $f(x,y)$ be a function of two variables. 
What is the counterexample that there exists $A$ s.t. for all $\theta$, $$\lim_{r\to 0+}f(r\cos \theta,r\sin \theta)=A$$ but double limit
$$
\lim_{(x,y)\to (0,0)}f(x,y)$$
does not exist?
Is it true:
$\lim_{(x,y)\to (0,0)}f(x,y)=A$  if and only if $$\lim_{r\to 0+}f(r\cos \theta,r\sin \theta)=A$$ uniformly for $\theta\in[0,2\pi)$?
 A: As you pointed out in your comment, the function
$$f(x,y)=
  \begin{cases}
    \frac{x^2y}{x^4+y^2} & (x,y)\ne(0,0), \\
    0 & x=y=0.
  \end{cases}
$$ 
is a counterexample.
If you substitute $x=r\cos\theta$, $y=r\sin\theta$, 
$$f(r,\theta)= \frac{r^3 \cos^2\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta}.$$
We have
$$f(r,\theta)= \frac{r^3 \cos^2\theta\sin\theta}{r^4\cos^4\theta+r^2\sin^2\theta} = 
\frac{r \cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta} \overset{(*)}=
\frac{r\sin\theta}{r^2\cos^2\theta+\tan^2\theta}.$$
For $r\to0^+$ and $\tan\theta\ne0$ the above expression tends to $0$. The cases $\sin\theta=0$ (which leads to $\tan\theta=0$) and $\cos\theta=0$ (which is not o.k. in the above argument, since that would mean that we divided by zero in the step marked $(*)$) can be discussed separately, but in both cases the limit is $0$.
This can be also said differently: If you tend to the origin along the line $y=kx$ where $k=\tan\theta$ you get
$$f(x,y)=\frac{k x^3}{x^4+k^2x^2} = \frac{kx}{x^2+k^2}.$$
For $k\ne 0$ we get limit equal to zero. The case $k=0$ can be dealt with separately. The expression $y=kx$ is symmetric to this one. (Fixing $\theta$ is the same as tending to zero along a line going through the origin.)

But the function given above does not converge to $0$ for $(x,y)\to(0,0)$. To see this, just notice that whenever $y=x^2$ and $x\ne0$, we get
$$f(x,y)=\frac{x^4}{2x^4}=\frac12.$$
So the value $1/2$ is attained arbitrarily close to the origin.
See also Prove that $\lim\limits_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}} = 0$

Now if you say that $f(r,\theta)$ converges to $A$ uniformly w.r.t. $\theta$, then this is precisely the definition of
$$\lim_{(x,y)\to(0,0)} f(x,y)=A.$$
The uniform convergence says that for any given $\varepsilon>0$ there exists $r_0>0$ such that for $0<r<r_0$ we have
$$|f(r,\theta)-A|<\varepsilon.$$
This is equivalent to saying that in the ball $B(0,r_0)$ around origin the values of the function $f$ are $\varepsilon$-close to $A$.
