Đáp án:
\(2)\dfrac{1}{{12}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 2x + 1}}{{\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x + 2} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \dfrac{{1 - 1}}{{\left( {1 + 2} \right)\left( {1 + \sqrt {2.1 - 1} } \right)}} = 0\\
2)\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} + 8 - 8}}{{{x^2}\left( {\sqrt[3]{{{{\left( {{x^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{x^2} + 8} \right)}} + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{{x^2}\left( {\sqrt[3]{{{{\left( {{x^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{x^2} + 8} \right)}} + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\sqrt[3]{{{{\left( {{x^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{x^2} + 8} \right)}} + 4}}\\
= \dfrac{1}{{\sqrt[3]{{{{\left( {{0^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{0^2} + 8} \right)}} + 4}} = \dfrac{1}{{12}}
\end{array}\)
