# Under what conditions can I interchange the order of limits for a function of two variable?

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
The above example is nice. Just to clarify, $f(x,y)$ should be defined as $0$ if either $x=0$ or $y=0$. Then $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$. But for $x\neq 0$, $\lim_{y\rightarrow 0} f(x,y)$ does not exist. – Michael Jun 28 '14 at 17:36

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.