If $x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$, then:
$\textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0  2\text{ but finite}$
We note that $x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1+x$. By the quadratic formula, $x=\dfrac{1\pm \sqrt{5}}{2}$. Because there are only positive square roots in $x$, $x$ must be positive, thus, it is $\dfrac{1+\sqrt{5}}{2}\approx 1.618$. Thus, $\boxed{\textbf{(C)}\ 1 < x < 2}$.