Đáp án:
\[\mathop {\lim }\limits_{x \to 2} \frac{{2 - \sqrt {x + 2} }}{{{x^2} - 3x + 2}} = - \frac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{2 - \sqrt {x + 2} }}{{{x^2} - 3x + 2}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{{2^2} - {{\sqrt {x + 2} }^2}}}{{2 + \sqrt {x + 2} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{2 - x}}{{2 + \sqrt {x + 2} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{\left( {2 + \sqrt {x + 2} } \right)\left( {x - 1} \right)}}\\
= \frac{{ - 1}}{{\left( {2 + \sqrt {2 + 2} } \right)\left( {2 - 1} \right)}}\\
= \frac{{ - 1}}{{4.1}} = - \frac{1}{4}
\end{array}\)
