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Suppose I have $f:\mathbb{R}^2 \to \mathbb{R}$. What conditions do I need to say that

$$\lim_{x \to a} \lim_{y \to b} f(x,y) = \lim_{y \to b} \lim_{x \to a} f(x,y)$$

?

What about in a more general case, by taking $X,Y$ and $Z$ topological (Hausdorff) spaces and $f$ from $X \times Y$ to $Z$ ?

Thank you

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It is a little hard to say, what are you after? E.g., it would be enough if you assume that the limit $L=\lim\limits_{(x,y) \to (a,b)} f(x,y)$ exists, then both of these limits are $L$. –  Lukas Geyer Oct 5 '12 at 19:23
    
For the function $$f ( x,\,y ) = \begin{cases} { x \sin { \frac{1} {y} } + y \sin{ \frac{1}{x} } , \quad x \ne 0, \;y \ne 0, {} \\ 0, \quad x=0, \;y=0 } \end{cases} $$ none of repeated limits $\lim\limits_{x \to 0} \lim\limits_{y \to 0} f(x,y), \quad \lim\limits_{y \to 0} \lim\limits_{x \to 0} f(x,y)$ does not exist, but double limit $$ \lim\limits_{x \to 0}_{y \to 0} f(x,y)=0.$$ –  M. Strochyk Oct 5 '12 at 20:38
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1 Answer

Theorem. Let $\{ F_t ; t\in T\}$ be a family of functions $F_t : X \rightarrow \mathbb{C}$ depending on a parameter t; let $\mathcal{B}_X$ be a base $X$ and $\mathcal{B}_{T}$ a base in $T$. If the family converges uniformly on $X$ over the base $\mathcal{B}_{T}$ to a function $F : X \rightarrow \mathbb{C}$ and the limit $\lim_{\mathcal{B}_{T}} F_t(x)=A_t$ exists for each $t\in T$, the both repeated limits $\lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))$ and $\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x))$ exist and the equality

$$ \lim_{\mathcal{B}_{X}}(\lim_{\mathcal{B}_{T}}F_t(x))=\lim_{\mathcal{B}_{T}}(\lim_{\mathcal{B}_{X}}F_t(x)) $$ holds.

This theorem can be found in books of Zorich (Mathematical Analysis II p. 381).

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