Biết lim(x^2+2012) căn bậc 7 (1-2x) - 2012/x = a/b , với a/b là phân số tối giản, a là số nguyên âm

\(\mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + 2012} \right)\sqrt[7]{{1 - 2x}} - 2012}}{x} = \mathop {\lim }\limits_{x \to 0} \left( {x\sqrt[7]{{1 - 2x}}} \right) + 2012\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[7]{{1 - 2x}} - 1} \right)}}{x}\)

\( = 2012\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[7]{{1 - 2x}} - 1}}{x}\)

Xét hàm số \(y = f(x) = \sqrt[7]{{1 - 2x}}\) ta có \(f\left( 0 \right) = 1\). Theo định nghĩa đạo hàm ta có:

\(f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[7]{{1 - 2x}} - 1}}{x}\);

\(f'\left( x \right) =  - \frac{2}{{7{{\left( {\sqrt[7]{{1 - 2x}}} \right)}^6}}} \Rightarrow f\left( 0 \right) =  - \frac{2}{7} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[7]{{1 - 2x}} - 1}}{x} =  - \frac{2}{7}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{x^2} + 2012} \right)\sqrt[7]{{1 - 2x}} - 2012}}{x} =  - \frac{{4024}}{7} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a =  - 4024}\\{b = 7}\end{array} \Rightarrow a + b =  - 4017} \right.\).

Đáp án: −4017.