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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

  • 0
    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 $L$.2012-10-05
  • 1
    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 $$ \lim\limits_{x \to 0}_{y \to 0} f(x,y)=0.$$2012-10-05
  • 0
    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.2014-06-28

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