A function $f(x):\mathbb R^n \rightarrow\mathbb R^1$ is continuous on $\mathbb R^n\setminus \{0\}$ and the limit $\lim_{x\rightarrow0} f(x)$ exists (in a sense is bounded) in any direction to origin.

Is this function continuous in the origin, or in order to be continuous there it has to have the same limit value there in all directions?

How to differentiate such two cases defining function continuity?

Same question arises concerning differentiability.

I appreciate any explanations.

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    $\begingroup$ No, to be continuous at the origin it must have the same limit in all directions. But even that is not enough. It also must have that same limit when approached in ways other than along straight lines. Some of the point of the $\varepsilon$-$\delta$ definition of continuity is to cover cases like this. $\endgroup$ – GEdgar Jul 10 '17 at 20:05
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    $\begingroup$ If "in all directions" means "along any curve" then it is true. If you restrict any curve to any ray then it is false. $\endgroup$ – kp9r4d Jul 10 '17 at 22:42

In order to be continuous, the limit must coincide in all directions and the function must assume that value at the origin.

However, even more is needed. The limit must coincide regardless of the path taken towards the origin (directions correspond to straight lines). It is poosible that the limit exist and coincide for all directions but fails to exist/coincide for some other path. A classical example is


which approaches $0$ along any line but is constant and equal to $\frac12$ along the curve $y=x^2$.

For differentiability, we have the same situation.

  • $\begingroup$ Thank you for the explanation. And what kind of "continuity" is the case when the limit is just limited (being say continues along any straight path to, of even through)? $\endgroup$ – MirOdin Jul 10 '17 at 20:15
  • $\begingroup$ Do you mean perhaps 'bounded near the origin'? Admittedly, this is weaker than saying that all directional limits exist, but I don't quite understand what you're asking. $\endgroup$ – Fimpellizieri Jul 10 '17 at 20:17
  • $\begingroup$ How to formally characterize a function which is $C^2$ continuous on $R^n \ 0$, is ($C^0$) continuous in origin and has bounded derivatives in any direction there (other than in same these words)? $\endgroup$ – MirOdin Jul 10 '17 at 20:23
  • $\begingroup$ Sorry for typo, I mean $R^n\setminus 0$ there in place of $R^n 0$. $\endgroup$ – MirOdin Jul 10 '17 at 20:36
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    $\begingroup$ Just checking: you mean $f\in C^2(\mathbb{R}^n\setminus\{0\})\cap C^0(\mathbb{R}^n)$ and $f$ has bounded directional derivatives near $0$? $\endgroup$ – Fimpellizieri Jul 10 '17 at 20:39

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