Are there any paths that will always show if there is a limit?
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.
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.
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.