Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{x - 1}}{{\sqrt {2 - x} - 1}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{\left( {\sqrt {2 - x} - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{2 - x - 1}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{1 - x}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \left( { - \sqrt {2 - x} - 1} \right)\\
= - \sqrt {2 - 1} - 1\\
= - 2\\
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( { - 2x} \right) = - 2\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right)\left( { = - 2} \right)
\end{array}$
Vậy hàm số liên tục tại x=1
