The value of the continued fraction $1+\frac{1}{1+\frac{1}{1+\cdots }}$ is

$\begin{array}{l}x=1+\frac{1}{1+\frac{1}{1+\cdots }}\\ x=1+\frac{1}{x}\\ x=\frac{x+1}{x}\\ x^2-x-1=0\\ x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\times 1\times -1}}{2\times 1}\\ =\frac{1\pm \sqrt{1+4}}{2}\\ x=\frac{1\pm \sqrt{5}}{2}\\ x=\frac{1+\sqrt{5}}{2}(\text{Since 'x'} \text{must be pos}\end{array}$