if want to check a function like the following for continuity, we can argue, that we only have to check the point $(0,0)$ since for all other points, it is a composition of continuouse functions.
$f(x,y)=\begin{cases}0, & (x,y)=(0,0) \\ \frac{x^2-y^2}{\sqrt{x^2+y^2}}, & (x,y)\neq (0,0) \end{cases}$
So what we want is: $\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{\sqrt{x^2+y^2}} = 0$
Which is kind of hard in eucledian coordinates since there are infinite possibilities to let $(x,y) \to (0,0)$.
So we switch to polarcoordinates. We then let $r\to 0$ and check if the resulting limit depends on $\varphi$. If it does, we have a limit which depends on the direction we are approaching it, thus, we don't have continuity, if it doesn't, the function is continous in $(0,0)$ and thus in all points.
So far, so good. Let's consider another example. (I basically made it up, so I hope it "works". Since I can't really check it I don't know what point the following function should take in $(1,1)$ if it would be continous)
$g(x,y)=\begin{cases}0, & (x,y)=(1,1) \\ \frac{x^2-y^2}{\sqrt{(x-1)^2+(y-1)^2}}, & (x,y)\neq (1,1) \end{cases}$
Anyway, assuming we want to check that function for continuitiy, what would we do?
My though is: Since we would like to do $\lim_{(x,y)\to(1,1)} g(x,y) = g(1,1)$ we would need the following in polarcoordinates:
$r\cos(\varphi)\overset{!}{=}1$
$r\sin(\varphi)\overset{!}{=}1$
This is only the case for $r=1$ or respective $\varphi$'s. So could I just change to polar coordiantes and check:
$\lim_{r\to 1} g(r,\varphi)$
We have $|f(x,y)-f(0,0)| \le \frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}$. This gives
$f(x,y) \to f(0,0)$ as $(x,y) \to (0,0)$.