Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} - \sqrt[3]{{7x + 1}}}}{{x - 1}} = - \frac{1}{{12}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x + 2} - \sqrt[3]{{7x + 1}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x + 2} - 2} \right) + \left( {2 - \sqrt[3]{{7x + 1}}} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {2x + 2} \right) - {2^2}}}{{\sqrt {2x + 2} + 2}} + \frac{{{2^3} - \left( {7x + 1} \right)}}{{{2^2} + 2.\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{2\left( {x - 1} \right)}}{{\sqrt {2x + 2} + 2}} - \frac{{7\left( {x - 1} \right)}}{{{{\sqrt[3]{{7x + 1}}}^2} + 2\sqrt[3]{{7x + 1}} + 4}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{2}{{\sqrt {2x + 2} + 2}} - \frac{7}{{{{\sqrt[3]{{7x + 1}}}^2} + 2.\sqrt[3]{{7x + 1}} + 4}}} \right]\\
= \frac{2}{{\sqrt {2.1 + 2} + 2}} - \frac{7}{{{{\sqrt[3]{{7.1 + 1}}}^2} + 2.\sqrt[3]{{7.1 + 1}} + 4}}\\
= \frac{2}{4} - \frac{7}{{12}}\\
= - \frac{1}{{12}}
\end{array}\)
