# Prove that for a function repeated Limit exist but simultaneous Limit does Not exist.

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

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

Also

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

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 $$0$$.

• 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!
– user435638
Sep 28, 2019 at 11:30
• @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. 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 $$x=y=t \to 0$$

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