Show that for the function defined by:

$f(x,y) = \begin{cases} 1 & xy =0\\ 0 & xy \ne 0\\ \end{cases}$

repeated limit exist at the origin but simultaneous limit does Not exist.

Now for repeated limit I can say that

$\displaystyle \lim_{x \to 0}( \lim_{y \to 0} f(x,y)) = \displaystyle \lim_{x \to 0}1 = 1$


$\displaystyle \lim_{y \to 0}( \lim_{x \to 0} f(x,y)) = \displaystyle \lim_{y \to 0}1 = 1$ so both limit exist and are equal.

But I am not sure how to prove the second part ie Simultaneous limit do not Exist

Can anyone help me in this case ?

Thank you.


2 Answers 2


It follows from what you did that if the limit $\lim_{(x,y)\to(0,0)}f(x,y)$ existed, then it would have to be $0$. But you also have $\bigl(\forall x\in(0,\infty)\bigr):f(x,0)=1$ and therefore the limit $\lim_{(x,y)\to(0,0)}f(x,y)$ cannot be $0$.

  • $\begingroup$ I just read your answer and realized I had calculated the repeated Limits incorrect, both repeated Limits are actually $1$ and Not $0$. But then also your reasoning is correct , that the simultaneous limits do Not exist. Thanks for the answer! $\endgroup$
    – user435638
    Sep 28, 2019 at 11:30
  • $\begingroup$ @user435638 and for others reading this, Just a remark, the repeated limits are indeed equal, but they are 0, not 1 as written by the questioner. $\endgroup$
    – user166305
    Aug 6, 2022 at 19:11

We have that for $y=0$

$$\lim_{(x,y)\to (0,0)} f(x,y)=\lim_{(x,0)\to (0,0)} f(x,y)=1$$

but for $x=y=t \to 0$

$$\lim_{(x,y)\to (0,0)} f(x,y)=\lim_{t\to 0} f(t,t)=0$$


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