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?

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, $(1)=\lim_{x\to0}1=1$.
