I got a piecewise function defined as $$f(x,y)= \begin{cases} 1, & 0<y<x^2\\ 0, &\quad\; \;\text{else} \end{cases}$$
How to find $\lim_{(x,y) \to (0,0)}f(x,y)$ along the $y=mx$ path ?
Besides, why does $n$ needed to bigger than or equal to zero for the following limit exist : $$\lim_{x \to a}g(x)^n=\lim_{ x\to a}[g(x)]^n$$
Thank you so much
For your $\lim_{(x,y) \to (0,0)} f(x,y)$ let's first consider this area. Here's a sketch of the region where $0<y<x^2$.
So we know that $f(x,y)$ is $1$ in the shaded region and it's $0$ everywhere else. Keep in mind, the area is $0<y<x^2$ so any point that lies on the line $y=0$ or $y=x^2$ will NOT be in this region, and our function will take the value of $0$ at these points.
What we want to do is think about the value of $f(x,y)$ for $(x,y)$ pairs very close to $(0,0)$ that lie on the $y=mx$ path. For any $m \neq 0$, there will be an $x \neq 0$ small enough so that $mx>x^2$ when $mx$ is positive. Clearly $mx$ could be negative if either $m$ or $x$ is negative. Since $m$ and $x$ are non-zero, we can then say that for $x$ values that are very close to $0$, our path $y=mx$ will always give us $mx<0$ or $mx>x^2$.
If $mx$ is always greater than $x^2$ or less than $0$ for $x$ values near $0$, then we can also say $y$ is always greater than $x^2$ or less than $0$ for $x$ values near $0$ (since $y=mx$).
This all assumed that $m \neq 0$. If instead $m=0$, then our path would be $y=0$. We already determined that $y=0$ is not in the shaded region.
Therefore, the path $y=mx$ does not lie in the shaded region for $x$ values very close to $0$. And for any $m$, if $x$ is very close to $0$, then so is $y$ when we're on this path. All of this means that $f(x,y)=0$ for all $(x,y)$ pairs very close to $(0,0)$ that also lay on the path $y=mx$.
You will also notice that as $x$ approaches $0$ and continues to get smaller, all of the above inequalities hold up. All this means is that $f(x,y)$ will continue to take on the value of $0$ as $(x,y) \to (0,0)$. Therefore, $$\lim_{(x,y) \to (0,0)}f(x,y)=0.$$