Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{2 - \sqrt {2x - 1} .\sqrt[3]{{5x + 3}}}}{{x - 1}} = - \frac{{29}}{{12}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt {2x - 1} .\sqrt[3]{{5x + 3}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt {2x - 1} .\left[ {\left( {\sqrt[3]{{5x + 3}} - 2} \right) + 2} \right]}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{2.\left( {1 - \sqrt {2x - 1} } \right) - \sqrt {2x - 1} \left( {\sqrt[3]{{5x + 3}} - 2} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{2.\frac{{{1^2} - \left( {2x - 1} \right)}}{{1 + \sqrt {2x - 1} }} - \sqrt {2x - 1} .\frac{{\left( {5x + 3} \right) - {3^3}}}{{{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + {2^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{2.\frac{{2\left( {1 - x} \right)}}{{1 + \sqrt {2x - 1} }} - \sqrt {2x - 1} .\frac{{5\left( {x - 1} \right)}}{{{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + 4}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\dfrac{{ - 4}}{{1 + \sqrt {2x - 1} }} - \dfrac{{5\sqrt {2x - 1} }}{{{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + 4}}} \right]\\
= \frac{{ - 4}}{{1 + \sqrt {2.1 - 1} }} - \frac{{5.\sqrt {2.1 - 1} }}{{{{\sqrt[3]{{5.1 + 3}}}^2} + 2.\sqrt[3]{{5.1 + 3}} + 4}}\\
= \frac{{ - 4}}{2} - \frac{5}{{12}}\\
= - \frac{{29}}{{12}}
\end{array}\)
