Đáp án:
\[\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{x - 11}} + \sqrt[4]{{2x + 10}}}}{{x - 3}} = \frac{7}{{48}}\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{x - 11}} + \sqrt[4]{{2x + 10}}}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {\sqrt[3]{{x - 11}} + 2} \right) + \left( {\sqrt[4]{{2x + 10}} - 2} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{x - 11 + {2^3}}}{{{{\sqrt[3]{{x - 11}}}^2} - 2.\sqrt[3]{{x - 11}} + 4}} + \frac{{\sqrt {2x + 10} - 4}}{{\sqrt[4]{{2x + 10}} + 2}}}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{x - 3}}{{{{\sqrt[3]{{x - 11}}}^2} - 2.\sqrt[3]{{x - 11}} + 4}} + \frac{{2x + 10 - 16}}{{\left( {\sqrt[4]{{2x + 10}} + 2} \right)\left( {\sqrt {2x + 10} + 4} \right)}}}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \left[ {\frac{1}{{{{\sqrt[3]{{x - 11}}}^2} - 2.\sqrt[3]{{x - 11}} + 4}} + \frac{2}{{\left( {\sqrt[4]{{2x + 10}} + 2} \right)\left( {\sqrt {2x + 10} + 4} \right)}}} \right]\\
= \frac{1}{{{{\sqrt[3]{{3 - 11}}}^2} - 2.\sqrt[3]{{3 - 11}} + 4}} + \frac{2}{{\left( {\sqrt[4]{{2.3 + 10}} + 2} \right)\left( {\sqrt {2.3 + 10} + 4} \right)}}\\
= \frac{1}{{12}} + \frac{2}{{32}}\\
= \frac{7}{{48}}
\end{array}\)
