# Weaking the path test for multivariable limits

The multivariable limit $$\lim_{(x,y)\to (x_0,y_0)} f(x,y)$$ exists and is equal to some scalar $$L$$ if and only if the limit $$\lim_{t\to 1} f(r(t))$$ exists and is equal to $$L$$ for all functions $$r:\mathbb{R}\to \mathbb{R}^2$$ such that $$\lim_{t\to 1} r(t) = (x_0, y_0)$$. It is common to use this fact as a way to conclude that a certain limit does not exist, by checking that taking the limit along different paths leads to different limits or by showing that the limit along a certain path does not exist. I wonder if this characterizarion can be weakened in the following way:

The multivariable limit $$\lim_{(x,y)\to (x_0,y_0)} f(x,y)$$ exists and is equal to some scalar $$L$$ if and only if $$\lim_{y\to y_0} f(x_0,y)=L$$ and $$\lim_{x\to x_0} f(x,g(x)) = L$$ for all functions $$g:\mathbb{R}\to \mathbb{R}$$ such that $$\lim_{x\to x_0} g(x) = y_0$$.

That is, it the limit exists and is the same going through all graphs of functions $$g(x)$$ and through the $$y$$ axis, can I conclude the existence of the multivariable limit?

EDIT: I substituted my initial condition of $$\lim_{y\to y_0} \lim_{x\to x_0} f(x,y)=L$$ by $$\lim_{x\to x_0} f(x,g(x)) = L$$ since, as José Carlos Santos noted in the comments, formalizes better the idea of approaching the limit vertically.

• It seems to me that what expresses the idea that the limit going through the $y$ axis is $L$ is $\lim_{y\to y_0}f(x_0,y)=L$. Sep 12, 2021 at 16:08
• @JoséCarlosSantos Yeah, I was unsure between the two conditions, but I think $lim_{y\to y_0} f(x_0, y) = L$ may be a better choice. I'll edit the question. Sep 12, 2021 at 17:04
• Related Sep 12, 2021 at 18:09
• @GiuseppeNegro Thanks, this is interesting. Makes me think the answer to my question is "no", though I can't think of a counterexample yet. Sep 12, 2021 at 19:24

Yes, you can.

Suppose that, for some $$L\in\Bbb R$$, you don't have $$\lim_{(x,y)\to(x_0,y_0)}f(x,y)=L$$. Then there is some $$\varepsilon>0$$ such that, for every $$\delta>0$$, there is some $$(x,y)\in D_f$$ such that$$\|(x,y)-(x_0,y_0)\|<\delta\quad\text{and that}\quad|f(x,y)-f(x_0,y_0)|\geqslant\varepsilon.$$In particular, for every $$n\in\Bbb N$$, there is some $$(x_n,y_n)\in D_f$$ such that$$\|(x_n,y_n)-(x_0,y_0)\|<\frac1n\quad\text{and that}\quad|f(x_n,y_n)-L|\geqslant\varepsilon.$$At least one of the following assertions holds:

1. We have $$x_n=x_0$$ for infinitely many $$n$$'s.
2. We have $$x_n>x_0$$ for infinitely many $$n$$'s.
3. We have $$x_n for infinitely many $$n$$'s.

If the first assertion holds, we can assume without loss of generality that $$x_n=x_0$$ for each $$n\in\Bbb N$$. But then, since $$\lim_{n\to\infty}(x_n,y_n)=(x_0,y_0)$$, $$\lim_{n\to\infty}y_n=y_0$$, and so we cannot have $$\lim_{y\to y_0}f(x_0,y)=L$$ (because $$|f(x_0,y_n)-L|\geqslant\varepsilon$$ for each $$n\in\Bbb N$$).

If the second assertion holds, we can assume without loss of generality that the sequence $$(x_n)_{n\in\Bbb N}$$ is strictly decreasing. Consider the only function $$g\colon[x_0,x_1]\longrightarrow\Bbb R$$ such that:

• $$(\forall n\in\Bbb Z_+):g(x_n)=y_n$$;
• the restriction of $$f$$ to each interval $$[x_{n-1},x_n]$$ is affine.

Then, since $$\lim_{n\to\infty}x_n=x_0$$, $$g$$ is continuous. But, since $$(\forall n\in\Bbb N):\bigl|f(x_n,g(x_n))-L\bigr|\geqslant\varepsilon$$, we cannot have $$\lim_{x\to x_0}f(x,g(x))=L$$.

What can be done under the third assumption is similar to what was done with the second one.

• Thank you, your argument is neat and easy fo follow. Sep 14, 2021 at 15:54
• I'm glad I could help. Sep 14, 2021 at 15:54