There's a thorem that say:

Let $f(x, y)$ be a function. if: $\lim_{y \to 0} [\lim_{x \to 0} f(x, y)] \ne \lim_{x \to 0} [\lim_{y \to 0} f(x, y)] \implies \text{ no limit }$

Is that theorem works in the opposite direction?

I mean,

Let $L = \lim_{y \to 0} [\lim_{x \to 0} f(x, y)] = \lim_{x \to 0} [\lim_{y \to 0} f(x, y)]$.

It implies that $\lim_{(x, y) \to (0, 0)} f(x, y) = L$?

Thanks in advance!

  • $\begingroup$ No, just take $f(x,y) = x/y$. The middle limit exists and is $0$, but the second one doesn't. $\endgroup$
    – IAmNoOne
    Feb 14, 2014 at 20:37
  • 1
    $\begingroup$ Try $f(x,y) = xy/(x^2+y^2)$. $\endgroup$ Feb 14, 2014 at 20:38

1 Answer 1


As it has also been mentioned in the comments you might look at

$f(x,y)=\frac{x^2y}{x^{4}+y^2}$ you can see $\lim_{y \to 0} [\lim_{x \to 0} f(x, y)] = \lim_{x \to 0} [\lim_{y \to 0} f(x, y)]=0$

but if $ (x,y)\implies (0,0)$ along $y=kx^2$ then different values of $k$ will give different values to the limit, and hence it does not exist.


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