(1) If 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) $\mathrm{C}$ Hint: Since $a>0$, then $a+1>1, x^{2}<1+a$ is equivalent to $x^{2} \leqslant 1, -1 \leqslant x \leqslant 1$.