Đáp án:
\[\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} }}{{\left( {x - 1} \right)\sqrt {{x^2} + 2} }} = - \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} }}{{\left( {x - 1} \right)\sqrt {{x^2} + 2} }}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} }}{{ - \left( {1 - x} \right)\sqrt {{x^2} + 2} }}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{ - 1}}{{\sqrt {1 - x} .\sqrt {{x^2} + 2} }}\\
\mathop {\lim }\limits_{x \to {1^ - }} \sqrt {1 - x} = \sqrt {1 - 1} = 0\\
\mathop {\lim }\limits_{x \to {1^ - }} \sqrt {{x^2} + 2} = \sqrt {{1^2} + 2} = \sqrt 3 \\
\sqrt {1 - x} .\sqrt {{x^2} + 2} \ge 0,\,\,\,\forall x \le 1\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} \sqrt {1 - x} .\sqrt {{x^2} + 2} = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} \frac{{ - 1}}{{\sqrt {1 - x} .\sqrt {{x^2} + 2} }} = - \infty
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} }}{{\left( {x - 1} \right)\sqrt {{x^2} + 2} }} = - \infty \)
