Đáp án:
7) \(\dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\mathop {\lim }\limits_{x \to 2} \dfrac{{4x - 8}}{{\left( {x - 2} \right)\left( {\sqrt[3]{{{{\left( {4x} \right)}^2}}} + 2\sqrt[3]{{4x}} + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{4}{{\sqrt[3]{{{{\left( {4x} \right)}^2}}} + 2\sqrt[3]{{4x}} + 4}}\\
= \dfrac{4}{{\sqrt[3]{{{{\left( {4.2} \right)}^2}}} + 2\sqrt[3]{{4.2}} + 4}} = \dfrac{1}{3}\\
7)\mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt x + 1}}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}}\\
= \dfrac{{\sqrt 1 + 1}}{{\sqrt[3]{{{1^2}}} + \sqrt[3]{1} + 1}} = \dfrac{2}{3}
\end{array}\)
