Đáp án:

k) \( - \dfrac{7}{{12}}\) 

Giải thích các bước giải:

\(\begin{array}{l}
c)\mathop {\lim }\limits_{x \to 2} \dfrac{{4 - x - 2}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {2 + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{2 - x}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {2 + \sqrt {x + 2} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - 1}}{{\left( {x - 1} \right)\left( {2 + \sqrt {x + 2} } \right)}} =  - \dfrac{1}{{\left( {2 - 1} \right)\left( {2 + \sqrt {2 + 2} } \right)}} =  - \dfrac{1}{4}\\
d)\mathop {\lim }\limits_{x \to 2} \dfrac{{4x + 1 - 9}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{4x - 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{4}{{\left( {x + 2} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}\\
 = \dfrac{4}{{\left( {2 + 2} \right)\left( {\sqrt {4.2 + 1}  + 3} \right)}} = \dfrac{1}{6}\\
e)\mathop {\lim }\limits_{x \to 1} \dfrac{{2x + 7 - 9}}{{\left( {x - 1} \right)\left( {{x^2} - 3x + 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{2x - 2}}{{\left( {x - 1} \right)\left( {{x^2} - 3x + 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{2}{{\left( {{x^2} - 3x + 3} \right)\left( {\sqrt {2x + 7}  + 3} \right)}}\\
 = \dfrac{2}{{\left( {{1^2} - 3.1 + 3} \right)\left( {\sqrt {2.1 + 7}  + 3} \right)}} = \dfrac{1}{3}\\
f)\mathop {\lim }\limits_{x \to 4} \dfrac{{x + 5 - 2x - 1}}{{\left( {x - 4} \right)\left( {\sqrt {x + 5}  + \sqrt {2x + 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 4} \dfrac{{ - x + 4}}{{\left( {x - 4} \right)\left( {\sqrt {x + 5}  + \sqrt {2x + 1} } \right)}}\\
 = \mathop {\lim }\limits_{x \to 4} \dfrac{{ - 1}}{{\sqrt {x + 5}  + \sqrt {2x + 1} }}\\
 = \dfrac{{ - 1}}{{\sqrt {4 + 5}  + \sqrt {2.1 + 1} }} =  - \dfrac{1}{6}\\
j)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + 1}}{{\left( {x + 1} \right)\left( {2x + 3} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\left( {2x + 3} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}}\\
 = \dfrac{1}{{\left( {2.1 + 3} \right)\left( {\sqrt[3]{{{1^2}}} + \sqrt[3]{1} + 1} \right)}} = \dfrac{1}{{15}}\\
k)\mathop {\lim }\limits_{x \to  - 2} \dfrac{{2x + 12 + {x^3}}}{{x\left( {x + 2} \right)\left( {\sqrt[3]{{{{\left( {2x + 12} \right)}^2}}} - x\sqrt[3]{{2x + 12}} + {x^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\left( {x + 2} \right)\left( {{x^2} - 2x + 6} \right)}}{{x\left( {x + 2} \right)\left( {\sqrt[3]{{{{\left( {2x + 12} \right)}^2}}} - x\sqrt[3]{{2x + 12}} + {x^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{{x^2} - 2x + 6}}{{x\left( {\sqrt[3]{{{{\left( {2x + 12} \right)}^2}}} - x\sqrt[3]{{2x + 12}} + {x^2}} \right)}}\\
 = \dfrac{{{{\left( { - 2} \right)}^2} - 2\left( { - 2} \right) + 6}}{{\left( { - 2} \right)\left( {\sqrt[3]{{{{\left( {2\left( { - 2} \right) + 12} \right)}^2}}} + 2\sqrt[3]{{2\left( { - 2} \right) + 12}} + {{\left( { - 2} \right)}^2}} \right)}}\\
 =  - \dfrac{7}{{12}}
\end{array}\)