Câu hỏi 49

Ta có: \(\mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{\sqrt {{x^3} + 2{x^2} + x + 4} {\rm{\;}} - 2}}{{x + 1}}\).

                                \(\begin{array}{l} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^3} + 2{x^2} + x + 4 - 4}}{{\left( {x + 1} \right)\left( {\sqrt {{x^3} + 2{x^2} + x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{x{{\left( {x + 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {\sqrt {{x^3} + 2{x^2} + x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^3} + 2{x^2} + x + 4}  + 2}} = 0 = f\left( { - 1} \right)\end{array}\)

\( \Rightarrow \) Hàm số đã cho liên tục tại \(x =  - 1\).

Ta có:

\(f'\left( 1 \right) = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{\sqrt {{x^3} + 2{x^2} + x + 4} {\rm{\;}} - 2}}{{{{\left( {x + 1} \right)}^2}}}\)

            \(\begin{array}{l} = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{{x^3} + 2{x^2} + x + 4 - 4}}{{{{\left( {x + 1} \right)}^2}\left( {\sqrt {{x^3} + 2{x^2} + x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 1} \dfrac{{x{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}\left( {\sqrt {{x^3} + 2{x^2} + x + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to  - 1} \dfrac{x}{{\sqrt {{x^3} + 2{x^2} + x + 4}  + 2}} = \dfrac{{ - 1}}{4}\end{array}\)

Chọn A.