Let $f(x)=x+\frac{1}{2 x+\frac{1}{2 x+\frac{1}{2 x+} \ddots}}$ for $x>0$. Find $f(99) f^{\prime}(99)$.
Solution: Assume that the continued fraction converges (it does) so that $f(x)$ is well defined. Notice that $f(x)-x=\frac{1}{x+f(x)}$, so $f(x)^{2}-x^{2}=1$, or $f(x)=\sqrt{1+x^{2}}$ (we need the positive square root since $x>0$ ). Thus $f^{\prime}(x)=\frac{x}{\sqrt{1+x^{2}}}$, so $f(x) f^{\prime}(x)=x$. In particular, $f(99) f^{\prime}(99)=\mathbf{9 9}$.