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Is there an $\mathbb{R}^2 \to \mathbb{R}$ function which does not have a limit at a point $a$ ($\nexists \lim\limits_{a} f$), but we get the same value as a possible value for the limit when we try to approach $f(a)$ through any line?

I think we would need $$f to not be differentiable around $a$. In that case, it would not be approximated by a linear function, so it could have a different limit with a non-linear path. However, I can't go further. And I don't even know if it's a goof approach.

This is not a homework, it was just mentioned by my calculus teacher in the previous semester.

Botond
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1 Answers1

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Sure, let $A = \{(r \cos t , r \sin t) | t \in (0,2 \pi] , 0 \le r \le t\}$.

Define $f = 1_A$, then $f$ does not have a limit as $x \to 0$ but has the limit $1$ along each line through the origin.

copper.hat
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  • It might be obvious, but how can you see that it does not have limit at $0$? – Botond Apr 17 '18 at 15:54
  • (It is easiest to see if you draw a picture of $A$.) If we let $x_n={2 \over n} (\cos {1 \over n}, \sin {1 \over n})$ then since $x_n \notin A$ we have $f(x_n) = 0$ and $x_n \to 0$. However with $y_n={1 \over n} (\cos {1 \over n}, \sin {1 \over n})$ we have $f(y_n) = 1$, and $y_n \to 0$. Hence $f$ cannot have a limit at $0$. – copper.hat Apr 17 '18 at 16:05
  • However, along each line through the origin, there is a small interval containing $0$ in its interior) on which $f$ is $1$, hence it has the limit $1$ along any path to $0$ (on the line). – copper.hat Apr 17 '18 at 16:06
  • It's possibly simpler to let $P={(x,x^2): x\in \mathbb R}$ and $f= \chi_P.$ – zhw. Apr 17 '18 at 16:20
  • Thanks, I see it now. Do you know any good software to draw a picture of A? – Botond Apr 17 '18 at 16:21
  • @zhw. has a simpler answer. – copper.hat Apr 17 '18 at 16:21
  • @zhw. What's $\chi_P$? – Botond Apr 17 '18 at 16:22
  • Another notation for the indicator function. – copper.hat Apr 17 '18 at 16:22
  • I like the shape of $A$, sort of a Pac-Man with under bite. – copper.hat Apr 17 '18 at 16:25
  • For @zhw. 's function, is it good to check it with the $x_n=\frac{1}{n}(\cos(t), \sin(t))$, for a fixed $t$, observe that $\cos(t)^2 \neq \sin(t)$, so $f(x_n) \to 0$, but for $x_n=(1/n, 1/n^2)$, $f(x_n) \to 1$? – Botond Apr 17 '18 at 16:42