Ta có \(f\left( x \right) = \left| x \right| = \left\{ \begin{array}{l}
x\,\,\,\,khi\,\,x \ge 0\\
x\,\,\,\,khi\,\,x < 0
\end{array} \right.\)

\(\begin{array}{l}
+ f\left( 0 \right) = 0\\
\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} x = 0;\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0
\end{array}\)

Do \(f\left( 0 \right) = 0 = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\) nên hàm số liên tục tại \(x_0=0\)

\(\begin{array}{l}
+\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \frac{x}{x} = 1 \Rightarrow f'\left( {{0^ + }} \right) = 1\\
\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \frac{{ - x}}{x} =  - 1 \Rightarrow f'\left( {{0^ - }} \right) =  - 1
\end{array}\)

Do \(f'\left( {{0^ + }} \right) \ne f'\left( {{0^ - }} \right)\) nên hàm số không tồn tại đạo hàm tại \(x_0=0\)