Phương pháp giải:

+) Hàm số liên tục tại \(x={{x}_{0}}\Leftrightarrow \underset{x\to x_{0}^{+}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to x_{0}^{-}}{\mathop{\lim }}\,f\left( x \right)=f\left( 0 \right)\)

+) Hàm số có đạo hàm tại \(x={{x}_{0}}\Leftrightarrow f'\left( {{x}_{0}} \right)=\underset{x\to x_{0}^{+}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}=\underset{x\to x_{0}^{-}}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}\)

Lời giải chi tiết:

Ta có: \(y = \frac{{\left| x \right|}}{{x + 1}} = \left\{ \begin{array}{l}\frac{x}{{x + 1}}\,\,\,khi\,x \ge 0\\\frac{{ - x}}{{x + 1}}\,\,\,khi\,\,x < 0\end{array} \right.\)

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

\(f'\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\frac{f\left( x \right)-f\left( 0 \right)}{x-0}\) 

Ta có:

\(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{x}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{x + 1}} = 1\\\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{\frac{{ - x}}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{ - 1}}{{x + 1}} =  - 1\end{array} \right. \Rightarrow x = {x_0} \Rightarrow \mathop {\lim }\limits_{x \to x_0^ + } \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} \ne \mathop {\lim }\limits_{x \to x_0^ - } \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} \Rightarrow \) Hàm số không tồn tại đạo hàm tại x = 0.

Chọn B.