repeated limit confusion Consider the below function.
$$f(x,y) = 
\begin{cases}
1 & xy \neq 0 \\
0 & xy = 0
\end{cases}$$
Suppose i want to calculate the below repeated limit.
$$\lim_{x\to0}\lim_{y\to0}f(x,y)$$.
In general textbooks in India for BS courses,the above repeated limit computation is given to be $1$ which is fine if we know that the point $(x,y)\neq 0$. However , if we be really precise, then ,this limit computation is$$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$$.
The inner limit is unknown to us as we donot know about the nature of $x$. So my question is , in repeated limit calculation , do we assume that there is some distance left that we still have to move some in $x$ and $y$ to approach a given point?
 A: The limit $\lim_{y\to0}f(x,y)$ is $0$ when $x=0$ and $1$ otherwise. But when you compute$$\lim_{x\to0}\left(\lim_{y\to0}f(x,y)\right),\tag1$$what happens when $x=0$ does not matter. So, $(1)=\lim_{x\to0}1=1$.
A: $$\lim_{x\to0}[\lim_{y\to0}f(x,y)]$$

The inner limit is unknown to us
as we donot know about the nature of $x$.

This is false. We do know the nature of $x$: it is either approaching $0$ from above or from below.
To calculate the double limit $$\large{\lim_{x\to0^-}[\lim_{y\to0}f(x,y)]},$$
fix $x<0$ and then calculate $\large{L_x = \lim_{y\to0}f(x,y).}\quad (1)$
The value of $\large{L_x}$ depends on the value of $x$ you fixed before you calculated $\lim_{y\to0}f(x,y).$
Now repeat process $(1)$ whilst taking values of $x<0$ closer and closer to $0.$
In other words,
$$\large{\lim_{x\to0^-}[\lim_{y\to0}f(x,y)] = \lim_{x\to0^-}L_x}.$$
$$$$
Similarly, To calculate the double limit $$\large{\lim_{x\to0^+}[\lim_{y\to0}f(x,y)]},$$
fix $x>0$ and then calculate $\large{L_x = \lim_{y\to0}f(x,y).}\quad (2)$
The value of $\large{L_x}$ depends on the value of $x$ you fixed before you calculated $\lim_{y\to0}f(x,y).$
Now repeat process $(2)$ whilst taking values of $x>0$ closer and closer to $0.$
In other words,
$$\large{\lim_{x\to0^+}[\lim_{y\to0}f(x,y)] = \lim_{x\to0^+}L_x}.$$
If $\large{\lim_{x\to 0^+}L_x}$ and  $\large{\lim_{x\to 0^+}L_x}$ both exist and are equal, then by definition they are equal to $\lim_{x\to0}[\lim_{y\to0}f(x,y)]$.
