# Limit existence of a multi-variable function

Having the same limit along all straight lines approaching $(x_0,y_0)$ does not imply a limit exists at $(x_0, y_0)$

Consider, $$\lim_{(x, y) \rightarrow (0,0)}f(x, y) = \frac{2x^2y}{x^4+y^2}$$

This function has a limit 0 along every path $y=mx$. However, the limit fails to exist as we move along the paths $y=kx^2$. In summary, the limit does not exist.

Suppose I come across a function whose limit exists and is some $L$ along two different sets of paths $f(x)$ and $g(x)$. Will this agreement be enough for me to conclude that the limit exists and is $L$?

There could be a third set of paths defined by the function $h(x)$ for which the limit fails to exist. How do I know for sure that a limit exists without having to check for several paths?

• Your questions asks for a whole lecture on evaluating limits. I think that it is too broad. Commented Dec 8, 2017 at 16:05

Paths are a good tool for investigating convergence but not for proving convergence. This is due to the nature of the definition: We say that $\lim_{(x, y) \to (0, 0)} f(x) = L$ if $$\forall \epsilon > 0, \exists \delta > 0, \|(x, y)\| < \delta \implies |f(x) - L| < \epsilon.$$ This definition has absolutely no reference to paths, but only to closeness to the point where we are taking the limit.
This is also one of the reasons that shifting to polar coordinates is frequently useful: Polar coordinates have a built-in parameter ($r$) to measure distance to the origin.
• But why $(x,y)=(x_0,y_0)$ is allowed? Commented Mar 20, 2018 at 7:45
Once you feel confident that the limit exist then try to use the $\delta-\epsilon$ definition of limit to prove the limit does exist and is the one that you are claiming it to be.