1. Let $f_{1}(x)=\frac{x}{\sqrt{1+x^{2}}}$, for any natural number $n$, define $f_{x+1}(x)=f_{1}\left(f_{n}(x)\right)$. Then the analytical expression for $f_{1993}(x)$ is ).
(A) $\frac{1993 x}{\sqrt{1+x^{2}}}$
(B) $\frac{x}{\sqrt{1993+x^{2}}}$
(C) $\frac{x}{\sqrt{1+1993 x^{2}}}$
(D) $\frac{1993 x}{\sqrt{1+1993 x^{2}}}$
,$- 1 . C$.
By induction, we can see that $f_{n}(x)=\frac{x}{\sqrt{1+n x^{2}}}$. When $n=1$, $f_{1}(x)=\frac{x}{\sqrt{1+x^{2}}}$. Assuming that when $n=k$, $f_{k}(x)=\frac{x}{\sqrt{1+k x^{2}}}$. Then, when $n=k+1$,
$$
\begin{aligned}
f_{k+1}(x) & =\frac{f_{k}(x)}{\sqrt{1+f_{k}^{2}(x)}}=\frac{\frac{x}{\sqrt{1+k x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+k x^{2}}}} \\
& =\frac{x}{\sqrt{1+(k+1) x^{2}}} .
\end{aligned}
$$

Therefore, $f_{1993}(x)=\frac{x}{\sqrt{1+1993 x^{2}}}$.