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Is the function $f(\theta,r) = \sin(\frac{r}{\theta}), f(0,0) = 0$ continous at the point $(0,0)$ To clarify, $\theta$ and $r$ are the vector $(\theta,r)$'s polar coordinates. Someone told me that the answer to my question is yes, but it looks like there is a series $(x_n,y_n)$ that tends to $0$ s.t. $\theta = r $, contradicting continuity. Any help would be appreciated!

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Indeed, the function as you've defined it at the point $(x,y) = (0,0)$ – Omnomnomnom Mar 8 '14 at 20:55
up vote 1 down vote accepted

No, $\sin(r/\theta)$ is not continuous. The function $f(x, y) = r \sin(\theta)$ is continuous, however -- perhaps there was a miscommunication.

(To prove it's not continuous, pick $\theta = \pi/2$ and $r_k = 1/(4k+1)$. Then $\theta_k / r_k$ is $\pi/2$ (mod $2 \pi$), so $f$ is one at all these points, but $0$ at their limit.

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I'm looking for a function that has directional derivatives at the point $(0,0)$ that are not bounded (each directional derivative is a finite number). This function seems to fit what I'm looking for, except for the continuity part. – ILoveLev Mar 8 '14 at 20:46
Whoops, my function was wrong, I edited it now. – ILoveLev Mar 8 '14 at 20:47
And now your function looks a little better, but it's not defined for $\theta = 0$. Perhaps you need to think about this a little longer. – John Hughes Mar 8 '14 at 20:49
Indeed, a function defined in terms of polar coordinates' $\theta$ should have the property that it's invariant under adding a multiple of $2\pi$ to $\theta$; your $\sin(r/\theta)$ is not, hence isn't really well-defined. – Greg Martin Mar 8 '14 at 21:22

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