jmc

We evaluate $r$ several times to see if there is a pattern. Indeed, $r(\theta )=\frac{1}{1-\theta }$, so $$\begin{array}{rl}r(r(\theta ))& =r\left(\frac{1}{1-\theta }\right)=\frac{1}{1-\frac{1}{1-\theta }}\cdot \frac{1-\theta }{1-\theta }\\ & =\frac{1-\theta }{1-\theta -1}=\frac{1-\theta }{-\theta }=1-\frac{1}{\theta }.\end{array}$$ Then, $$r(r(r(\theta )))=r\left(1-\frac{1}{\theta }\right)=\frac{1}{1-\left(1-\frac{1}{\theta }\right)}=\frac{1}{\frac{1}{\theta }}=\theta .$$ Hence, for any $\theta$, we have that $r(r(r(\theta )))=\theta$ is the identity. Then, $$r(r(r(r(r(r(30))))))=r(r(r(30)))=\boxed{30}.$$