Đáp án:

$\begin{array}{l}
9)\mathop {\lim }\limits_{x \to 8} \dfrac{{x - 8}}{{\sqrt[3]{x} - 2}}\\
 = \mathop {\lim }\limits_{x \to 8} \dfrac{{\left( {\sqrt[3]{x} - 2} \right)\left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)}}{{\sqrt[3]{x} - 2}}\\
 = \mathop {\lim }\limits_{x \to 8} \left( {\sqrt[3]{{{x^2}}} + 2\sqrt[3]{x} + 4} \right)\\
 = \left( {\sqrt[3]{{{8^2}}} + 2\sqrt[3]{8} + 4} \right)\\
 = 12\\
10)\mathop {\lim }\limits_{x \to 4} \dfrac{{3 - \sqrt {5 + x} }}{{1 - \sqrt {5 - x} }}\\
 = \mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {3 - \sqrt {5 + x} } \right)\left( {3 + \sqrt {5 + x} } \right)\left( {1 + \sqrt {5 - x} } \right)}}{{\left( {1 - \sqrt {5 + x} } \right)\left( {1 + \sqrt {5 - x} } \right)\left( {3 + \sqrt {5 + x} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {9 - 5 - x} \right)\left( {1 + \sqrt {5 - x} } \right)}}{{\left( {1 - 5 + x} \right)\left( {3 + \sqrt {5 + x} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 4} \dfrac{{ - \left( {1 + \sqrt {5 - x} } \right)}}{{3 + \sqrt {5 + x} }}\\
 = \dfrac{{ - \left( {1 + \sqrt {5 - 4} } \right)}}{{3 + \sqrt {5 + 4} }}\\
 = \dfrac{{ - 1}}{3}\\
11)\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3}  - \sqrt {3x + 1} }}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {x + 3}  - 2 + 2 - \sqrt {3x + 1} }}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{x + 3 - 4}}{{\sqrt {x + 3}  + 2}} + \dfrac{{4 - 3x - 1}}{{2 + \sqrt {3x + 1} }}}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{1}{{\sqrt {x + 3}  + 2}} - \dfrac{3}{{2 + \sqrt {3x + 1} }}} \right)\\
 = \dfrac{1}{{\sqrt {1 + 3}  + 2}} - \dfrac{3}{{2 + \sqrt {3.1 + 1} }}\\
 = \dfrac{1}{4} - \dfrac{3}{4} = \dfrac{{ - 1}}{2}\\
12)\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1}  - 3}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - x - 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{\left( {4x + 1 - 9} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 1} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4.\left( {x - 2} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4\left( {x + \sqrt {x + 2} } \right)}}\\
 = \dfrac{{\left( {2 + 1} \right)\left( {\sqrt {4.2 + 1}  + 3} \right)}}{{4.\left( {2 + \sqrt {2 + 2} } \right)}}\\
 = \dfrac{{18}}{{4.4}} = \dfrac{9}{8}\\
13)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{x\sqrt x  + 1}}{{{x^2} + x + 1}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\dfrac{{x\sqrt x }}{{{x^2}}} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\dfrac{1}{{\sqrt x }} + \dfrac{1}{{{x^2}}}}}{{1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}}} = 0
\end{array}$