Xét hai câu sau:

(1) Hàm số \(y = \dfrac{{\left| x \right|}}{{x + 1}}\) liên tục tại $x = 0.$

(2) Hàm số \(y = \dfrac{{\left| x \right|}}{{x + 1}}\) có đạo hàm tại $x = 0.$

Trong 2 câu trên:

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

Ta có $\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{x}{{x + 1}} = 0 = f\left( 0 \right)\\\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{ - 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 $x = 0.$

\(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}\) 

Ta có:

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