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I am curious, if there can be a function $f(x,y)$, which is continuous in a point $[0,0]$, but for which iterated limit $\lim _{x \to 0} \lim _{y \to 0} (f(x,y))$ does not exist.

Is it even possible for the function to be continuos without those limits?

Can you give me an example of this function?

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    $\begingroup$ Did you mean $\lim _{x \to 0} f(x,0)$ and $\lim _{y \to 0} f(0,y)$? $\endgroup$ – zhw. Apr 20 '15 at 18:18
  • $\begingroup$ I should improve it a little bit. I would rather use the iterated limit $\lim_{x \to 0} \lim_{y \to 0} f(x,y)$ $\endgroup$ – PObdr Apr 20 '15 at 18:41
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If $\lim_{y\to 0}f(x,y)$ exists for $x\ne 0,$ then the iterated limit will exist and equal $f(0,0).$ But it's possible for $f$ to be continuous at $(0,0)$ and not have that first limit existing: Let $g$ be the characteristic function of the rationals on $\mathbb {R}.$ Set $f(x,y)=x^2g(y).$ Then $f$ is continuous at $(0,0)$ but for $x\ne 0,\lim_{y\to 0}f(x,y)$ does not exist.

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No, it's not possible by definition of continuous function:

f(x,y) is continuous at point $(0,0)$ iff for all $\varepsilon>0$ exists $\delta>0$ that:

$$\sqrt{h_1^2+h_2^2}<\delta \Rightarrow \left|f(h_1,h_2)-f(0,0)\right|<\varepsilon$$

We can put $h_1=\delta$ and $h_2=0$, that means that

$$\lim_{x \to 0}f(x,0)=f(0,0)$$

The same with the other limit.

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