题目

设(x)=x(x-1)(x-2)... (x-99), 则-|||-'(0)= ()()-|||-A、99-|||-B、 -99-|||-C、99!-|||-D、 -991!.

$\begin{aligned} &\text{设}f\left(x\right)=x\left(x-1\right)\left(x-2\right)\cdots(x-99)\text{,则} \\ &f^{\prime}\left(0\right)=\left(\right) \\ &\begin{array}{cc}A\text{、}&99\end{array} \\ &B\text{、}-99 \\ &\text{C、99!} \\ &D\text{、}-99! \end{aligned}$ .

题目解答

答案

解：

$\begin{gathered} f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0} \\ =\lim_{x\to0}\frac{x(x-1)\cdots(x-99)-0}{x-0} \\ =\lim_{x\to0}(x-1)(x-2)\cdots(x-99) \\ =(-1)(-2)\cdots(-99)=-99! \end{gathered}$

所以，最终答案，选择D.