smc

We note that $x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots }}}=1+x$. By the quadratic formula, $x=\frac{1\pm \sqrt{5}}{2}$. Because there are only positive square roots in $x$, $x$ must be positive, thus, it is $\frac{1+\sqrt{5}}{2}\approx 1.618$. Thus, $\boxed{1<x<2}$.