1. Given that the real number $x$ satisfies for any positive number $a$, $x^{2}<1+a$. Then the minimum value of $x$ is ( ).
(A) 0
(B) 1
(C) -1
(D) Does not exist:
-, 1. C.
From $a>0 \Rightarrow a+1>1$.
Then $x^{2}<1+a \Leftrightarrow x^{2} \leqslant 1$.
Therefore $-1 \leqslant x \leqslant 1$.