Giải thích các bước giải:

\(\begin{array}{l}
{C_5} = \mathop {\lim }\limits_{x \to  - 2} \frac{{\sqrt {x + 11}  - \sqrt[3]{{8x + 43}}}}{{2{x^2} + 3x - 2}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {\sqrt {x + 11}  - 3} \right) + \left( {3 - \sqrt[3]{{8x + 43}}} \right)}}{{\left( {x + 2} \right)\left( {2x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\frac{{\left( {x + 11} \right) - {3^2}}}{{\sqrt {x + 11}  + 3}} + \frac{{{3^3} - \left( {8x + 43} \right)}}{{{3^2} + 3.\sqrt[3]{{8x + 43}} + {{\sqrt[3]{{8x + 43}}}^2}}}}}{{\left( {x + 2} \right)\left( {2x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\frac{{x + 2}}{{\sqrt {x + 11}  + 3}} - 8.\frac{{x + 2}}{{9 + 3\sqrt[3]{{8x + 43}} + {{\sqrt[3]{{8x + 43}}}^2}}}}}{{\left( {x + 2} \right)\left( {2x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \frac{{\frac{1}{{\sqrt {x + 11}  + 3}} - \frac{8}{{9 + 3\sqrt[3]{{8x + 43}} + {{\sqrt[3]{{8x + 43}}}^2}}}}}{{2x - 1}}\\
 = \dfrac{{\frac{1}{{\sqrt { - 2 + 11}  + 3}} - \frac{8}{{9 + 3.\sqrt[3]{{8.\left( { - 2} \right) + 43}} + {{\sqrt[3]{{8.\left( { - 2} \right) + 43}}}^2}}}}}{{2.\left( { - 2} \right) - 1}}\\
 = \dfrac{{\frac{1}{6} - \frac{8}{{27}}}}{{ - 5}}\\
 = \frac{7}{{270}}\\
{C_7} = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{8x + 11}} - \sqrt {x + 7} }}{{{x^2} - 3x + 2}}\\
 = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt[3]{{8x + 11}} - 3} \right) + \left( {3 - \sqrt {x + 7} } \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\frac{{\left( {8x + 11} \right) - {3^3}}}{{{{\sqrt[3]{{8x + 11}}}^2} + \sqrt[3]{{8x + 11}}.3 + {3^2}}} + \frac{{{3^2} - \left( {x + 7} \right)}}{{3 + \sqrt {x + 7} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\frac{{8\left( {x - 2} \right)}}{{{{\sqrt[3]{{8x + 11}}}^2} + 3\sqrt[3]{{8x + 11}} + 9}} - \frac{{x - 2}}{{3 + \sqrt {x + 7} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\frac{8}{{{{\sqrt[3]{{8x + 11}}}^2} + 3\sqrt[3]{{8x + 11}} + 9}} - \frac{1}{{3 + \sqrt {x + 7} }}}}{{x - 1}}\\
 = \dfrac{{\frac{8}{{{{\sqrt[3]{{8.2 + 11}}}^2} + 3.\sqrt[3]{{8.2 + 11}} + 9}} - \frac{1}{{3 + \sqrt {2 + 7} }}}}{{2 - 1}}\\
 = \frac{8}{{27}} - \frac{1}{6} = \frac{7}{{54}}\\
{C_9} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{7 + {x^3}}} - \sqrt {3 + {x^2}} }}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt[3]{{7 + {x^3}}} - 2} \right) + \left( {2 - \sqrt {3 + {x^2}} } \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {7 + {x^3}} \right) - {2^3}}}{{{{\sqrt[3]{{7 + {x^3}}}}^2} + 2\sqrt[3]{{7 + {x^3}}} + 4}} + \frac{{{2^2} - \left( {3 + {x^2}} \right)}}{{2 + \sqrt {3 + {x^2}} }}}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{{{\sqrt[3]{{7 + {x^3}}}}^2} + 2\sqrt[3]{{7 + {x^3}}} + 4}} - \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{2 + \sqrt {3 + {x^2}} }}}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{{x^2} + x + 1}}{{{{\sqrt[3]{{7 + {x^3}}}}^2} + 2\sqrt[3]{{7 + {x^3}}} + 4}} - \frac{{x + 1}}{{2 + \sqrt {3 + {x^2}} }}} \right]\\
 = \frac{{{1^2} + 1 + 1}}{{{{\sqrt[3]{{7 + {1^3}}}}^2} + 2.\sqrt[3]{{7 + {1^3}}} + 4}} - \frac{{1 + 1}}{{2 + \sqrt {3 + {1^2}} }}\\
 = \frac{3}{{12}} - \frac{2}{4}\\
 =  - \frac{1}{4}
\end{array}\)