Câu hỏi 27

\(\lim L = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{{x - 2}} - \frac{1}{{{x^2} - 4}}} \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{{x + 2 - 1}}{{{x^2} - 4}}} \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{{x + 1}}{{{x^2} - 4}}} \right) =  - \infty \)

(Vì \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 1} \right) = 3 > 0;\,\,\mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 4} \right) = 0;\,\,x \to {2^ - } \Rightarrow {x^2} - 4 < 0\))

Chọn D.