Đáp án:
\[\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - \sqrt {x + 2} }}{{\sqrt[4]{{2x + 2}} - 2}} = - \frac{8}{{27}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt[3]{{4x - 1}} - \sqrt {x + 2} }}{{\sqrt[4]{{2x + 2}} - 2}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{\left( {\sqrt[3]{{4x - 1}} - 3} \right) + \left( {3 - \sqrt {x + 2} } \right)}}{{\sqrt[4]{{2x + 2}} - 2}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{\frac{{4x - 1 - {3^3}}}{{{{\sqrt[3]{{4x - 1}}}^2} + 3.\sqrt[3]{{4x - 1}} + 9}} + \frac{{{3^2} - x - 2}}{{3 + \sqrt {x + 2} }}}}{{\frac{{\sqrt {2x + 2} - 4}}{{\sqrt[4]{{2x + 2}} + 2}}}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{4.\frac{{x - 7}}{{{{\sqrt[3]{{4x - 1}}}^2} + 3.\sqrt[3]{{4x - 1}} + 9}} - \frac{{x - 7}}{{3 + \sqrt {x + 2} }}}}{{\frac{{2x - 14}}{{\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt {2x + 2} + 4} \right)}}}}\\
= \mathop {\lim }\limits_{x \to 7} \left[ {\left( {\frac{4}{{{{\sqrt[3]{{4x - 1}}}^2} + 3.\sqrt[3]{{4x - 1}} + 9}} - \frac{1}{{3 + \sqrt {x + 2} }}} \right):\frac{2}{{\left( {\sqrt[4]{{2x + 2}} + 2} \right)\left( {\sqrt {2x + 2} + 4} \right)}}} \right]\\
= \left[ {\left( {\frac{4}{{{{\sqrt[3]{{4.7 - 1}}}^2} + 3.\sqrt[3]{{4.7 - 1}} + 9}} - \frac{1}{{3 + \sqrt {7 + 2} }}} \right):\frac{2}{{\left( {\sqrt[4]{{2.7 + 2}} + 2} \right)\left( {\sqrt {2.7 + 2} + 4} \right)}}} \right]\\
= \left[ {\left( {\frac{4}{{27}} - \frac{1}{6}} \right):\frac{2}{{32}}} \right] = - \frac{8}{{27}}
\end{array}\)
