Đáp án:
\(VD7:\dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
VD4:\\
\mathop {\lim }\limits_{x \to 0} \dfrac{{2\left( {3x + 1 - 1} \right)}}{{x\left( {\sqrt {3x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{6x}}{{x\left( {\sqrt {3x + 1} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{6}{{\sqrt {3x + 1} + 1}} = \dfrac{6}{{\sqrt {3.0 + 1} + 1}} = \dfrac{6}{3} = 2\\
VD5:\\
\mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {x - 3} \right)\left( {\sqrt {4x + 1} + 1} \right)}}{{4x + 1 - 1}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {x - 3} \right)\left( {\sqrt {4x + 1} + 1} \right)}}{{4x}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {x - 3} \right)\left( {\sqrt {4x + 1} + 1} \right)}}{4}\\
= \dfrac{{\left( {0 - 3} \right)\left( {\sqrt {4.0 + 1} + 1} \right)}}{4} = \dfrac{{ - 3.2}}{4} = - \dfrac{3}{2}\\
VD6:\\
\mathop {\lim }\limits_{x \to 5} \dfrac{{\left( {3x + 1 - 16} \right)\left( {3 + \sqrt {x + 4} } \right)}}{{\left( {\sqrt {3x + 1} + 4} \right)\left( {9 - x - 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 5} \dfrac{{\left( {3x - 15} \right)\left( {3 + \sqrt {x + 4} } \right)}}{{\left( {\sqrt {3x + 1} + 4} \right)\left( {5 - x} \right)}}\\
= \mathop {\lim }\limits_{x \to 5} \dfrac{{ - 3\left( {3 + \sqrt {x + 4} } \right)}}{{\sqrt {3x + 1} + 4}}\\
= \dfrac{{ - 3\left( {3 + \sqrt {5 + 4} } \right)}}{{\sqrt {3.5 + 1} + 4}} = - \dfrac{{18}}{7}\\
VD7:\\
\mathop {\lim }\limits_{x \to - 2} \dfrac{{\left( {x + 1} \right) + 1}}{{\left( {x + 2} \right)\left( {\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} - \sqrt[3]{{\left( {x + 1} \right)}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{1}{{\sqrt[3]{{{{\left( {x + 1} \right)}^2}}} - \sqrt[3]{{\left( {x + 1} \right)}} + 1}}\\
= \dfrac{1}{{\sqrt[3]{{{{\left( { - 2 + 1} \right)}^2}}} - \sqrt[3]{{\left( { - 2 + 1} \right)}} + 1}} = \dfrac{1}{3}
\end{array}\)
