Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\left( {2x + 1} \right)\left( {3x + 1} \right)\left( {4x + 1} \right)} - 1}}{x} = \frac{9}{2}\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\left( {2x + 1} \right)\left( {3x + 1} \right)\left( {4x + 1} \right)} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\left( {6{x^2} + 5x + 1} \right)\left( {4x + 1} \right)} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {24{x^3} + 6{x^2} + 20{x^2} + 5x + 4x + 1} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {24{x^3} + 26{x^2} + 9x + 1} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{24{x^3} + 26{x^2} + 9x + 1 - 1}}{{\sqrt {24{x^3} + 26{x^2} + 9x + 1} + 1}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{24{x^3} + 26{x^2} + 9x}}{{\sqrt {24{x^3} + 26{x^2} + 9x + 1} + 1}}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{24{x^2} + 26x + 9}}{{\sqrt {24{x^3} + 26{x^2} + 9x + 1} + 1}}\\
= \frac{{{{24.0}^2} + 26.0 + 9}}{{\sqrt {{{24.0}^3} + {{26.0}^2} + 9.0 + 1} + 1}}\\
= \frac{9}{2}
\end{array}\)
