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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?

sjm23
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  • The first limit you gave is simply of the form ${f(x) + g(y)}$, where both $f$ and $g$ are continuous (in the regular single variable sense). it's a theorem that the multi variable limit is just the sum of the individual limits in this case – Riemann'sPointyNose Sep 11 '20 at 20:48
  • To prove harder limits you usually have to use multi variable limit properties, see here: https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/12%3A_Functions_of_Several_Variables/12.02%3A_Limits_and_Continuity_of_Multivariable_Functions to see the list of them – Riemann'sPointyNose Sep 11 '20 at 20:51
  • A limit can't be path dependent. If it is the limit itself does not exist. There is no straightforward easy way to tell when this is or is not the case. – K.defaoite Sep 11 '20 at 20:56
  • @K.defaoite I think the point here is that different paths may each separately have a well-defined limit (the limit along the path), but they may not have a common limit - the different paths may have different limits. The limits along the paths can exist in this case. – Mark Bennet Sep 11 '20 at 21:33
  • @Mark Bennet By the limit I meant the global limit, but I understand what you're saying. – K.defaoite Sep 11 '20 at 21:36

3 Answers3

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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.

Ninad Munshi
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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.

JimN
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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

user
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    But isn't that just a circle argument? in order for me to verify that it is continuous, at the point $(x_0,y_0)$ I should prove that the limit $\lim {(x, y) \rightarrow\left(x{0}, y_{0}\right)} f(x, y)=f\left(x_{0}, y_{0}\right)$ exist.

    In my book I'm supposed to solve these limits before learning about continuity, so how would I do that?

     – sjm23 Sep 11 '20 at 20:44
  • @sjm23 your first function is the sum of continuous single variable functions - it is a Theorem that the multi variable limit in this case is just the sum of the two individual limits – Riemann'sPointyNose Sep 11 '20 at 20:49
  • @sjm23 If the function is made by the composition of continuous "elementary funtions" and we can plug in the values and obtain a value then we can state that the function is continuous at that point. – user Sep 11 '20 at 20:49
  • @sjm23 You use the "heuristic" to get an answer (as it is just a heuristic at this stage, you may not have proven continuity yet in a sufficiently rigorous book), then you can justify it with $\epsilon-\delta$ if you wish. Scratch work does not need to be rigorous it is a place to build intuition. – Ninad Munshi Sep 11 '20 at 20:50