Suppose I wanted to show some multivariable (specifically, 2 variables, is what im referring to) function wasn't continuous. What ways are there to go about doing that? From what I know, there seem to be only 2 ways.
- Take the partial derivative of one of the variables and then let that approach 0 before dealing with the other variable.
- Finding some approach to 0, for which the function has more than one unique limit. e.g set limit as t-> infinity, x=t^-1 and y=t^-2 for which the limit =0 and x=t^-2 and y=t^-1 for which the limit goes to some c =/0
Have I missed out on some common approaches to these types of problems?
The most general approach is to define two sequences that converge to your desired point such that the sequence of function values (evaluated on these sequences) converge to two different values, or at least one of the limits of function values doesn't exist.
Often by inspecting your function, you can parametrize your two sequences, as in your $(t^{-1},t^{-2})$ and $(t^{-2},t^{-1})$ example, and just show you get two different limits. This is an art, coming up with the two parametrizations. Sometimes something simple like $(t^{-1},at^{-1})$ or $(at^{-1},t^{-1})$ for two different constants $a$ works. Also, sometimes converting to polar coordinates can help, where you realize how to define $\theta$ as two different functions of $r$ and then show you get different limits as $r \to 0$.
Keep in mind that for a "removable" discontinuity though, you may need one sequence of function values to be the single value of the function evaluated at the desired point.
Finally, another general approach: If you can find $\epsilon > 0$ such that for any little ball around your desired point, you can find two points whose function values are at least $\epsilon$ apart, this also shows the function is discontinuous. This is perhaps one of the most straightforward way to show for example that the function that is $f(x,y) = 1$ if $x,y$ are rational and zero otherwise is discontinuous everywhere.