Tính giới hạn \(F=\lim \limits_{x \rightarrow 0} \frac{\sqrt{(2 x+1)(3 x+1)(4 x+1)}-1}{x}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} F = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {(2x + 1)(3x + 1)(4x + 1)} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {} - 1} \right)\left( {\sqrt {(2x + 1)(3x + 1)(4x + 1)} + 1} \right)}}{{x\left( {\sqrt {(2x + 1)(3x + 1)(4x + 1)} + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{(2x + 1)(3x + 1)(4x + 1) - 1}}{{x\left( {\sqrt {(2x + 1)(3x + 1)(4x + 1)} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{24{x^3} + 26{x^2} + 9x}}{{x\left( {\sqrt {(2x + 1)(3x + 1)(4x + 1)} + 1} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{24{x^2} + 26x + 9}}{{\sqrt {(2x + 1)(3x + 1)(4x + 1)} + 1}} = \frac{9}{2} \end{array}\)