Are there any paths that will always show if there is a limit? I'm trying to do limits in 3D and I'm wondering whether or not there are paths along which the limit of any function at any point can always be found. In my book it isn't clear whether this exists or not; neither is it clear how to choose a path if this does exist.
In the book they replace $y$ with $kx$ a lot but sometimes they replace $x$ and $y$ with $0$ separately  and sometimes they use $x^2$.
I've seen some people have already asked similar questions but about specific formulas and I can't link that to my exact question.
Thanks for any help!
 A: I know what you're asking, and the answer is "no" (caveat after reading the comments: There are paths that do conclude for you, most importantly so-called space-filling curves, but they are mostly of theoretical value, and rarely help in calculations). Checking different (simple) paths is not enough to show that a limit exists. It is good for two things, though:


*

*Finding a candidate limit. If you pick a path, and calculate the limit along that path, then you have a candidate for what the limit could be. Most limit calculations become easier once you have a concrete candidate to check.

*Showing that there is no limit. If two different paths create two different candidates, or if a single pay fails to give a limit, then there cannot be a single limit value at the given point.


The standard example is
$$
f(x,y)=\cases{0&if $x=y=0$\\\frac{2x^2y}{x^4+y^2}&otherwise}
$$for which every line through the origin says that the limit at the origin is $0$, while the parabola $y=x^2$ says that the limit is $1$. In general, there is no way of knowing that there isn't such a "different path". For instance, this example is not difficult to change into one where the path that gives $1$ is $y=x^{10}$ instead. How will you know in general that there isn't such a path hidden somewhere? That's impossible without proving the general limit in the first place.
A: Interesting nontrivial theorem (simplified version of teorema 5 in Límites utilizando coordenadas polares, La Gaceta de la RSME, v. 7, n. 2.):
Let be $f$ defined in a neighborhood of $(0,0)$, $F(r,\theta) = f(r\cos\theta,r\sin\theta)$. We have the following equivalence:
$$\lim_{(x,y)\to(0,0)}f(x,y) = L\iff \forall\bar{\theta}\in[0,2\pi] \lim_{(r,\theta)\to(0,\bar{\theta})}F(r,\theta) = L.$$
(the second limit isn't a directional limit)
The essential ingredient of the proof is that $[0,2\pi]$
 is compact.
A: Correct me if I'm wrong, but I believe existence of a limit may indeed be seen by only looking at a single path, if admittedly an odd path.
Let's say you want to know whether $f$ has a limit at $(0,0)\in\mathbb{R}^2$. Let $\gamma_1$ be a space-filling curve, i.e. a path whose image is the (entire!) square $[-1,1]^2$. Let $\gamma_n:=\gamma_1/n$, i.e. a path whose image is the square with side lengths $2/n$. Finally, let $\gamma$ be the concatenation of all those paths, with straight segments connecting the end of $\gamma_{n-1}$ with  the start of $\gamma_n$. You can do all of this in such a way that $\gamma$ is defined on $[0,1]$ with $\gamma(1)=(0,0)$
Now if $f$ does have a limit at $(0,0)$, then $f\circ \gamma$ will have the same limit, as $\gamma$ after some while will live in the square $[-1/n,1/n]$ only. Conversely, if $f\circ \gamma$ has a limit $a$, then so does $f$ at $(0,0)$ since if there existed $x_n\to (0,0)$, $f(x_n)\not\to a$ then $(f\circ\gamma)(y_n)\not\to a$, where $y_n$ are chosen such that $\gamma(y_n)=x_n$ and $y_n\to 1$ (which is possible by surjectivity of each $\gamma_n$ onto a neighborhood of $(0,0)$.)
The practicability of this approach to verify limits by hand is of course limited, but we are just mathematicians after all.
