$13 \cdot 16$ If $0<x<1$, then among $x^{2}, x, \sqrt{x}, \frac{1}{x}$
(A) $\frac{1}{x}$ is the largest, $x^{2}$ is the smallest.
(B) $x$ is the largest, $\frac{1}{x}$ is the smallest.
(C) $x^{2}$ is the largest, $\sqrt{x}$ is the smallest.
(D) $x$ is the largest, $x^{2}$ is the smallest.
(China Junior High School Mathematics Correspondence Competition, 1987)
[Solution] From the given condition $01 \text {. }$

Since $x-x^{2}=x(1-x)>0$, we have $x>x^{2}$. Also, $\sqrt{x}-x^{2}=\sqrt{x}(1-x \sqrt{x})>0$, so $\sqrt{x}>x^{2}$. Therefore, $\frac{1}{x}$ is the largest, and $x^{2}$ is the smallest. Hence, the answer is $(A)$.