Hàm nào trong các hàm số sau không có giới hạn tại điểm \(x = 2\)
Giải chi tiết:
Đáp án A ta có : \(\left. \matrix{ \mathop {\lim }\limits_{x \to {2^ + }} \left| {x - 2} \right| = \mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0 \hfill \cr \mathop {\lim }\limits_{x \to {2^ - }} \left| {x - 2} \right| = \mathop {\lim }\limits_{x \to {2^ - }} \left( { - x + 2} \right) = 0 \hfill \cr} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \left| {x - 2} \right| = \mathop {\lim }\limits_{x \to {2^ - }} \left| {x - 2} \right| = 0 \Rightarrow \mathop {\lim }\limits_{x \to 2} \left| {x - 2} \right| = 0\)
Đáp án B ta có : \(\left. \matrix{ \mathop {\lim }\limits_{x \to {2^ + }} {1 \over {\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ + }} {1 \over {x - 2}} = + \infty \hfill \cr \mathop {\lim }\limits_{x \to {2^ - }} {1 \over {\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ - }} {1 \over { - x + 2}} = + \infty \hfill \cr} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} {1 \over {\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ - }} {1 \over {\left| {x - 2} \right|}} = + \infty \Leftrightarrow \mathop {\lim }\limits_{x \to 2} {1 \over {\left| {x - 2} \right|}} = + \infty \)
Đáp án C ta có: \(\left. \matrix{ \mathop {\lim }\limits_{x \to {2^ + }} {1 \over {x - 2}} = + \infty \hfill \cr \mathop {\lim }\limits_{x \to {2^ - }} {1 \over {x - 2}} = - \infty \hfill \cr} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} {1 \over {\left| {x - 2} \right|}} \ne \mathop {\lim }\limits_{x \to {2^ - }} {1 \over {\left| {x - 2} \right|}} \Rightarrow \) Không tồn tại \(\mathop {\lim }\limits_{x \to 2} {1 \over {x - 2}}\)
Chọn C.