Cho hàm số \(f \left( x \right) = \left \{ \begin{array}{l}2 \left| x \right| - 1 \, \, \, \, \,khi \, \,x \le - 2 \ \ \sqrt {2{x^2} + 1} \, \,khi \, \,x > - 2 \end{array} \right. \). Tìm \( \mathop { \lim } \limits_{x \to {{ \left( { - 2} \right)}^ - }} f \left( x \right), \, \, \mathop { \lim } \limits_{x \to {{ \left( { - 2} \right)}^ + }} f \left( x \right) \) và \( \mathop { \lim } \limits_{x \to - 2} f \left( x \right) \) nếu có.
A.\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = 3\)
\(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = 3\).
B.\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = 2\)
\(\mathop {\lim }\limits_{x \to \left( { - 2} \right)} f\left( x \right) = 2\).
C.\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = 0\)\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = -1\)
D.\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} f\left( x \right) = + \infty\)
\(\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} f\left( x \right) = -1\)