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Can someone give me an example to show that in general

$\lim_{y \rightarrow b} \lim_{x \rightarrow a} f(x,y) \neq \lim_{(x,y)\rightarrow (a,b)} f(x,y) \neq \lim_{x \rightarrow a} \lim_{y \rightarrow b} f(x,y) $

I have been able to construct examples when the first and last limit exist but the middle one does not, but I can't find one where the middle limit exists but is not equal to the other two (both of which exist).

Does the existence of the middle limit as well as the other two imply that they must be equal to the middle one?

R_D
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    Use '\lim' instead of 'lim' to get $\lim_{x \to \infty}$ instead of $lim_{x \to \infty}$ – Krijn Aug 24 '15 at 06:11

3 Answers3

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The limit $$\lim_{x\to 0}\biggl(\lim_{y\to 0} x\sin\frac{1}{xy}\biggl)$$ does not exists. But $$\lim_{y\to 0}\biggl(\lim_{x\to 0} x\sin\frac{1}{xy}\biggl)=0$$

Mark
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    Thank you. But my question was if the non equality is true when all three limits exist. I modified it now for clarity. – R_D Aug 24 '15 at 06:39
  • Oh but I see that here $\lim_{(x,y)\rightarrow(0,0)} f = 0$ and is not equal to one of the other limits so this is the closest thing to the answer I need. Thank you! – R_D Aug 24 '15 at 06:45
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    Oh, ok. Excuse me. – Mark Aug 24 '15 at 06:48
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Now since $\lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y)$ and $\lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y)$ both exist it is obvious that $\lim_{x\rightarrow a} f(x,y)$ and $\lim_{y\rightarrow b} f(x,y)$ have to exist. For if not then the former two limits cannot be be calculated.

Lemma - If $\lim_{(x,y)\rightarrow (a,b)} f(x,y) = l$ and if both $\lim_{x\rightarrow a} f(x,y)$ and $\lim_{y\rightarrow b} f(x,y)$ exist then $\lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y) = l = \lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y)$.

Proof -

Define the functions $g_1(x) = \lim_{y\rightarrow b} f(x,y)$ and $g_2(y) = \lim_{x\rightarrow a} f(x,y)$.

Given an, $\epsilon > 0$

  1. There exists a $\delta_1 >0$ such that $|f(x,y) -l| < \epsilon /2$ whenever $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta_1$

  2. There exists a $\delta_2 > 0$ such that $|f(x,y) - g_1(x)| < \epsilon /2$ whenever $0 < |y - b| < \delta_2$

Let $\delta = \min\{\delta_1,\delta_2\}$

When both $0<|y-b|<\delta /\sqrt{2}$ and $0<|x-a|<\delta /\sqrt {2}$ we have $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$. So,

$$|g_1(x) - l| = |g_1(x) - f(x,y) + f(x,y) - l| \leq |f(x,y) - g_1(x)| + |f(x,y) - l| < \epsilon /2 + \epsilon /2 = \epsilon$$

Thus $\lim_{x\rightarrow a} g_1(x) = l$.

Similarly for $g_2(y)$. $\square$

Thus if all three limits in the question exist then they have to be equal.

R_D
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This is an example of a function that is discontinuous at $x=0=y$ and has different limits when that point is approached from different directions.

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Keith McClary
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  • Doesn't the definition say that whichever way you approach $(a,b)$ the limit exists and is the same? – Keith McClary Aug 25 '15 at 01:58
  • Yes. But for that you have to approach $(a,b)$ through a well defined path. If one of the inner limits doesn't exist then this path from which you want to approach $(a,b)$ is not defined at all. – R_D Aug 25 '15 at 06:35
  • @Rise As I read the definition the (two variable) limit does not exist by definition unless $f(x,y)$ becomes close to L for all $x,y$ close to $a,b$ . – Keith McClary Aug 25 '15 at 16:11