Suppose that we want to show that the function
$$f(z)=\begin{cases} \frac{z^5}{|z|^4},z\neq0\\ 0,z=0 \end{cases}$$
satisfies the Cauchy-Riemann equations at $z=0$. We see that
$$\frac{z^5}{|z|^4}=\frac{z^3}{\bar z^2}=\frac{x^3+3x^2iy-3xy^2-iy^3}{x^2-2ixy-y^2}$$
So finding $u$ and $v$ is slightly hard.
But the polar form of $\frac{z^5}{|z|^4}$ is $re^{5i\theta}=r\cos5\theta+ir\sin5\theta$. So
$$ru_r=r\cos5\theta,v_\theta=5r\cos5\theta\\ rv_r=r\sin5\theta,-u_\theta=5r\sin5\theta$$
But $\theta$ is undefined at $z=0$. So I want to know if we can put $r=0$ and conclude that
$$ru_r=v_\theta,rv_r=-u_\theta?$$
So at all what can we do here?
Note that$$u(x,y)=\operatorname{Re}\bigl(f(x+yi)\bigr)=\operatorname{Re}\left(\frac{(x+yi)^5}{|x+yi|^2}\right)$$(if $(x,y)\neq(0,0)$). Therefore, $x\neq0\implies u(x,0)=\operatorname{Re}x=x$ and so $\frac{\partial u}{\partial x}(0,0)=1$. On the other hand, $y\neq0\implies u(0,y)=\operatorname{Re}(yi)=0$ and so $\frac{\partial u}{\partial y}(0,0)=0$.
Using the same approach, it is easy to see that $\frac{\partial v}{\partial x}(0,0)=0$ and that $\frac{\partial v}{\partial y}(0,0)=1$. Therefore, $f$ satisfies the Cauchy-Riemann equations at $(0,0)$.