Giới hạn lim x → − 1 f 2 ( x ) + f ( x ) − 6 x + 1 bằng

C

Đặt \(g\left( x \right) = \frac{{f\left( x \right) - 2}}{{x + 1}}\)\( \Rightarrow f\left( x \right) = \left( {x + 1} \right)g\left( x \right) + 2\).

Ta cần tính \(\mathop {\lim }\limits_{x \to  - 1} f\left( x \right)\).

Ta có \(\mathop {\lim }\limits_{x \to  - 1} f\left( x \right)\)\( = \mathop {\lim }\limits_{x \to  - 1} \left[ {\left( {x + 1} \right)g\left( x \right) + 2} \right] = 2\).

Ta có \[\mathop {\lim }\limits_{x \to  - 1} \frac{{{f^2}\left( x \right) + f\left( x \right) - 6}}{{x + 1}}\]\[ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left[ {f\left( x \right) + 3} \right]\left[ {f\left( x \right) - 2} \right]}}{{x + 1}}\]

\[ = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left[ {f\left( x \right) - 2} \right]}}{{x + 1}}.\mathop {\lim }\limits_{x \to  - 1} \left[ {f\left( x \right) + 3} \right] = 2024.\left( {2 + 3} \right) = 2024.5 = 10120\].