\(\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + x} \right)\left( {1 + 2x} \right)\left( {1 + 3x} \right) - 1}}{x}\)
Giải chi tiết:
\(\mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + x} \right)\left( {1 + 2x} \right)\left( {1 + 3x} \right) - 1}}{x}\)
\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2{x^2} + 3x + 1} \right)\left( {1 + 3x} \right) - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{6{x^3} + 11{x^2} + 6x + 1 - 1}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {6{x^2} + 11x + 6} \right)}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {6{x^2} + 11x + 6} \right) = 6\end{array}\)
Chọn D.