Đáp án:
\[\frac{1}{{12}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} - \sqrt[3]{{7x + 1}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x + 2} - 2} \right) + \left( {2 - \sqrt[3]{{7x + 1}}} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2x + 2 - 4}}{{\sqrt {2x + 2} + 2}} + \frac{{8 - \left( {7x + 1} \right)}}{{4 + 2\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2\left( {x - 1} \right)}}{{\sqrt {2x + 2} + 2}} - \frac{{7\left( {x - 1} \right)}}{{4 + 2\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left( {\frac{2}{{\sqrt {2x + 2} + 1}} - \frac{7}{{4 + 2\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}} \right)\\
= \frac{2}{{\sqrt {2.1 + 2} + 1}} - \frac{7}{{4 + 2.2 + {2^2}}} = \frac{2}{3} - \frac{7}{{12}} = \frac{1}{{12}}
\end{array}\)
