\(\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{1 - x}}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{1 - x}}\)
\(\begin{array}{l}\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} 1 = 1 > 0\\\mathop {\lim }\limits_{x \to {1^ + }} \left( {1 - x} \right) = 1 - 1 = 0\\x > 1 \Rightarrow 1 - x < 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{1}{{1 - x}} = - \infty \\\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {1^ - }} 1 = 1 > 0\\\mathop {\lim }\limits_{x \to {1^ - }} \left( {1 - x} \right) = 1 - 1 = 0\\x < 1 \Rightarrow 1 - x > 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{1}{{1 - x}} = + \infty \end{array}\)
Vậy không tồn tại \(\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{1 - x}}.\)
Chọn D.