When is a multivariable limit path independent? When I had calculus I was taught that the limit of a multivariable limit can be path-dependent.
So In order to check if a limit exists, you should, in theory, check every possible path, which is infinitely many. So how do I actually calculate a multivariable limit? Just because I have checked one path, it doesn't necessarily mean the limit would be the same at every path?
Is there an easy way to know whether a limit is path independent, or when a multivariable limit might be path dependent?

Consider the limit:
$$\lim _{(x, y) \rightarrow(2,3)} 2x^3-y^{2}=16-9=7$$
How do I know that I can just put in the values in this case?
$$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2} y}{x^{4}+y^{2}}$$
I know this limit does not exist, because if you go along the path $y=mx$ the limit is 0. But if you go along the parabola $y=\pm x^2$ the limit is $\pm \frac{1}{2}$.
How are these two cases different. I mean how can you immediately see that the first case is path independent, but the second case may not be?
 A: Even if you don't have a formal definition of continuity just yet, you can take a look at the expression
$$\lim_{(x,y)\to(2,3)}2x^3-y^2$$
and think these functions look "nice" enough (we are not doing anything illegal like dividing by zero anywhere), so let's plug in the numbers and try to prove that for any $\epsilon > 0$ there exists $\delta > 0 $ such that
$$\sqrt{(x-2)^2+(y-3)^2} < \delta \implies |2x^3-y^2-7| < \epsilon$$
by re-centering our polynomial
$$2(x-2+2)^3-(y-3+3)^2 - 7$$
$$ = 2(x-2)^3+12(x-2)^2+24(x-2)+16-(y-3)^2-6(y-3)-9-7$$
$$ = 2(x-2)^3+12(x-2)^2+24(x-2)-(y-3)^2-6(y-3)$$
This means by triangle inequality (plus a domain restriction) we have that
$$|2x^3-y^2-7| < 38|x-2|+7|y-3| = 45\left[\sqrt{(x-2)^2+(y-3)^2}\right]$$
thus we can prove our limit by choosing
$$\delta = \min\left(1,45\epsilon\right)$$
The beauty of this is that this is an inequality without an appeal to continuity. We could do something similar with squeeze theorem. But anyway, later on, we come back and look at these functions where we were allowed to plug things in and get the limit anyway and formalize what made these special.
A: You are looking for continuity (but continuity in a multi-variable sense).
The first expression is just a polynomial, so there's no way for continuity to be interrupted. But the second one has a denominator, which comes with restrictions like when the denominator = 0.
A: In the first case the function is continuous at the point, since it is a composition of continuous elementary functions and it is defined at that point, that is
$$\lim _{(x, y) \rightarrow(x_0,y_0)} f(x,y)=f(x_0,y_0)$$
in this case limit always exists and it is therefore "path independent".
In the second case the function is not defined at the point and in this case the limit may not be "path independent".
Refer also to the related

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*How to prove that all elementary functions are continuous in their domain?
