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

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, $\epsilon > 0$


*

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

*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.

