Đáp án:
a) 5
Giải thích các bước giải:
\(\begin{array}{l}
a)\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2x + 1} \right)\left( {3x + 1} \right) - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{6{x^2} + 5x + 1 - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {6x + 5} \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left( {6x + 5} \right) = 5\\
b)\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {x + 1 - 1} \right)\left( {\sqrt {4x + 1} + 1} \right)}}{{\left( {4x + 1 - 1} \right)\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {\sqrt {4x + 1} + 1} \right)}}{{4x\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {4x + 1} + 1}}{{\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} + \sqrt[3]{{x + 1}} + 1}}\\
= \dfrac{{\sqrt {4.0 + 1} + 1}}{{\sqrt[3]{{{{\left( {0 + 1} \right)}^2}}} + \sqrt[3]{{0 + 1}} + 1}} = \dfrac{2}{3}
\end{array}\)
