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.
Best Answer
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 $0$.