How to show that this limit only exist when $\theta$=0, $\pi/2$, $\pi$, $3\pi/2$ $\pmod {2 \pi}$ Question
Consider the function $f(x,y) = (xy)^{1/3}$
Show that $\nabla f(0,0)=(0,0)$ and that the limit:
$$\lim_{t\to 0} \frac{f(t\cos(\theta),t\sin(\theta))}{t}$$
fails to exist when $\theta \neq 0,\pi/2, \pi, 2\pi/3$ $\pmod {2\pi}.$
Attempt
I get that $\nabla f =(\frac{y^{1/3}}{3x^{2/3}}, \frac{x^{1/3}}{3y^{2/3}})$ and that the limit is
$$\lim_{t\to 0}\frac{(t^2\cos(\theta) \sin(\theta))^{1/3} }{t}$$
but I don't know how to continue with it, I assume it has something to do with the $\nabla f(0,0)=(0,0)$ but $\nabla f(0,0)$ wouldn't exist if $\theta$=$0$
 A: Hint:
Observe that
$$\frac{f(t\cos\theta\,,\,\,t\sin\theta)}t=\frac{\left(\sin2\theta\right)^{1/3}}{2^{1/3}\,t^{1/3}}$$
The only way the above thing has a limit when $\;t\to0\;$ is if the numerator vanishes...
A: We consider the limit as t$\rightarrow 0$ for the function$\frac{\left(\sin2\theta\right)^{1/3}}{2^{1/3}\,t^{1/3}}$. We have two possibilities where the numerator is equal to $0$ or is not equal to $0$ and we will consider these in two cases separately below:
Case 1
If (sin $2\theta)^{1/3}$ is non-zero then as we take the limit the function diverges to infinity. Since we are dividing by a variable that approaches $0$.
That is because we know for $\alpha \neq 0$, that $\frac{\alpha}t$ converges to $\infty$ as $t\rightarrow 0$.
Case 2
However, we can resolve this problem by considering what happens when (sin $2\theta)^{1/3}$ = $0$. Now when we take the limit, when we take the limit we converge to $0$. In this case, the limit does exist.
Note: we can see this directly as when the numerator is equal to $0$, we see that
$\frac{\left(\sin2\theta\right)^{1/3}}{2^{1/3}\,t^{1/3}}$ = $\frac{0}{2^{1/3}\,t^{1/3}}=0$ and so it is clear that when $t \rightarrow  0$, this doesn't affect the fact that the function will always be equal to $0$, and so the limit is well defined at these points.
We now try to find the values of $\theta$ that make the numerator equal to $0$. We know that:
$$(sin2\theta)^{\frac{1}3}=0 \iff sin2\theta=0 \iff \theta=0,\frac{\pi}2,\pi, ...$$
Conclusion
The limit exists when $\theta=0,\frac{\pi}2,\pi, ...$ however, when theta is any other value, the limit does not exist as we end up diverging to infinity as t approaches $0$.
