Let $f(x,y)= \frac{xy}{x^2+y^2}$
prove that for all $z=(z_1,z_2) \neq(0,0)$, there exist $\lim_{(x,y)\rightarrow z}f(x,y)$ and find it.
My guess is that the result is $\frac{z_1z_2}{z_1^2+z_2^2}$. I wanted to convert the coordinates to modules and use the squeeze theorem, say, let $a=(x,y)$, then $|a|^2=x^2+y^2$, but I don't know how to relate the numerator to $|a|$.
In order to show that a limit in $\mathbb{R}^2$ exists, we must show that regardless of the direction from which we approach the point, the function approaches the same value.
In order to do this, we let the distance from the point go to 0 and see how the function behaves as a function of distance from the point.
We can express any point $(x,y)$ as $(x_0 + r\cos \theta, y_0 + r\sin \theta)$, where $r$ is the distance from the point $(x_0,y_0)$ and $\theta$ specifies the direction of the displacement from $(x_0,y_0)$. With this, we can say:
\begin{align} \lim_{(x,y)\to(x_0,y_0)} \frac{xy}{x^2+y^2} &= \lim_{r\to0} \frac{(x_0 + r\cos \theta)(y_0 + r\sin \theta)}{(x_0 + r\cos \theta)^2+(y_0 + r\sin \theta)^2} \\ &= \lim_{r\to0} \frac{x_0y_0 + r(y_0\cos \theta + x_0\sin \theta) + r^2 \cos \theta \sin \theta}{x_0^2 + y_0^2+2r (x_0\cos \theta + y_0\sin \theta) + r^2} \\ &= \frac{x_0y_0}{x_0^2 + y_0^2} \end{align} Note that we can say this specifically because $(x_0,y_0) \neq (0,0)$, so the denominator of the fraction in the limit does not tend to $0$.
If we were taking this limit to the origin instead, we would get \begin{align} \lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2} &= \lim_{r\to0} \frac{(r\cos \theta)(r\sin \theta)}{(r\cos \theta)^2+(r\sin \theta)^2} \\ &= \lim_{r\to0} \frac{r^2 \cos \theta \sin \theta}{r^2} \\ &= \lim_{r\to0} \cos \theta \sin \theta \\ &= \cos \theta \sin \theta \end{align} that is, the limit depends on what the direction from which we approach the origin, that is to say, the limit as we approach the origin does not exist.