Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\cos \dfrac{{\pi x}}{2}\,\,\,\,khi\,\,\left| x \right| \le 1\\\left| {x - 1} \right|\,\,\,\,\,\,\,khi\,\,\left| x \right| > 1\end{array} \right.\). Khẳng định nào sau đây đúng nhất?

\(f\left( x \right) = \left\{ \begin{array}{l}\cos \dfrac{{\pi x}}{2}\,\,\,\,khi\,\,\left| x \right| \le 1\\\left| {x - 1} \right|\,\,\,\,\,\,\,khi\,\,\left| x \right| > 1\end{array} \right. \Leftrightarrow f\left( x \right) = \left\{ \begin{array}{l}\cos \dfrac{{\pi x}}{2}\,\,\,\,khi\,\, - 1 \le x \le 1\\\left| {x - 1} \right|\,\,\,\,\,\,\,khi\,\,\left[ \begin{array}{l}x > 1\\x <  - 1\end{array} \right.\end{array} \right.\)

Ta có:

$\left. \begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left| {x - 1} \right| = 0\\\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \cos \dfrac{{\pi x}}{2} = \cos \dfrac{\pi }{2} = 0\\f\left( 1 \right) = \cos \dfrac{\pi }{2} = 0\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right) = 0 \Rightarrow $ Hàm số liên tục tại $x = 1.$

$\left. \begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \cos \dfrac{{\pi x}}{2} = \cos \dfrac{{ - \pi }}{2} = 0\\\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left| {x - 1} \right| = 2\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) \Rightarrow $Hàm số không liên tục tại \(x =  - 1\).