Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\dfrac{{\sqrt {2{x^2} + 8} - x - 2}}{{{x^2} - 4}}\,\,khi\,x < 2}\\{\,\,\,\,\,\dfrac{{\sin \left( {x - 2} \right)}}{{{x^2} - 3x + 2}}\,\,\,\,\,\,\,\,\,\,khi\,\,\,x > 2}\end{array}} \right.\). Tìm \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right),\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) nếu có.
A.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=1,\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=0\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) không tồn tại
B.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=0,\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=1\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) không tồn tại
C.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 2} f\left( x \right)=1\)
D.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 2} f\left( x \right)=0\)
A.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=1,\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=0\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) không tồn tại
B.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=0,\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=1\) và \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) không tồn tại
C.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 2} f\left( x \right)=1\)
D.\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)=\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)=\mathop {\lim }\limits_{x \to 2} f\left( x \right)=0\)
Loga
Hóa Học lớp 12
