Xét hai câu sau: (1) Hàm số y = | x | x + 1 liên tục tại x = 0. (2) Hàm số y = | x | x + 1 có đạo hàm tại x = 0. Trong 2 câu trên:

Ta có: \[{\rm{y = }}\frac{{\left| {\rm{x}} \right|}}{{{\rm{x + 1}}}} = \left\{ {\begin{array}{*{20}{c}}{\frac{x}{{x + 1}}\,\,khi\,\,x \ge 0}\\{\frac{{ - x}}{{x + 1}}\,\,khi\,\,x < 0}\end{array}} \right.\]

Ta có \(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{x + 1}} = 0 = f\left( 0 \right)}\\{\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - x}}{{x + 1}} = 0}\end{array}} \right. \Rightarrow \mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}^ - } f\left( x \right) = f\left( 0 \right) = 0 \Rightarrow \)Hàm số liên tục tại x = 0.

\[{\rm{f'}}\left( 0 \right) = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{f}}\left( {\rm{x}} \right) - {\rm{f}}\left( {\rm{0}} \right)}}{{{\rm{x}} - {\rm{0}}}}\]

Ta có:

\(\left\{ {\begin{array}{*{20}{c}}{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{x}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{x + 1}} = 1}\\{\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\frac{{ - x}}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 1}}{{x + 1}} = - 1}\end{array}} \right. \Rightarrow x = {x_0}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to {x_0}^ + } {\frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}_{}} \ne \mathop {\lim }\limits_{x \to {x_0}^ - } \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}}\)

Hàm số không tồn tại đạo hàm tại x = 0.

Đáp án cần chọn là: B