Đáp án:
\(3)\dfrac{1}{{12}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 3} \right)\left( {x - 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{\left( {4x + 1 - 9} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 3} \right)\left( {x - 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{4\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 3} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{4}\\
= \dfrac{{\left( {2 - 3} \right)\left( {\sqrt {4.2 + 1} + 3} \right)}}{4} = - \dfrac{3}{2}\\
2)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 2x + 1}}{{\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 1} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x + 2} \right)\left( {x + \sqrt {2x - 1} } \right)}}\\
= \dfrac{{1 - 1}}{{\left( {1 + 2} \right)\left( {1 + \sqrt {2.1 - 1} } \right)}} = 0\\
3)\mathop {\lim }\limits_{x \to 0} \dfrac{{8 + {x^2} - 8}}{{{x^2}\left( {\sqrt[3]{{{{\left( {{x^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{x^2} + 8} \right)}} + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{\sqrt[3]{{{{\left( {{x^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{x^2} + 8} \right)}} + 4}}\\
= \dfrac{1}{{\sqrt[3]{{{{\left( {{0^2} + 8} \right)}^2}}} + 2\sqrt[3]{{\left( {{0^2} + 8} \right)}} + 4}} = \dfrac{1}{{12}}
\end{array}\)
