Đáp án:

\(a.\mathop {\lim }\limits_{x \to  - 1} \frac{{2{x^2} + x + 1}}{{x - 2}} =  - \frac{2}{3}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to  - 1} \frac{{2{x^2} + x + 1}}{{x - 2}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{2{{\left( { - 1} \right)}^2} - 1 + 1}}{{ - 1 - 2}}\\
 =  - \frac{2}{3}\\
b.\mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {x - 1} \right)}}{{x - 2}}\\
 = \mathop {\lim }\limits_{x \to 2} \left( {x - 1} \right)\\
 = 2 - 1 = 1\\
c.\mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x - 1} \right){{\left( {x + 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{x - 2}}\\
 = \frac{{\left( { - 2 - 1} \right).0}}{{ - 4}} = 0\\
d.\mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {2x + 1} \right)\left( {x + 1} \right)\left( {\sqrt {{x^2} + 3}  + 2} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {2x + 1} \right)\left( {\sqrt {{x^2} + 3}  + 2} \right)}}{{x - 1}}\\
 = \frac{{\left( { - 2 + 1} \right)\left( {2 + 2} \right)}}{{ - 1 - 1}} = 2\\
e.\mathop {\lim }\limits_{x \to  - 3} \frac{{\left( {x + 12 - {x^2}} \right)\left( {1 + \sqrt {4 + x} } \right)}}{{\left( {\sqrt {x + 12}  - x} \right)\left( {1 - 4 - x} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 3} \frac{{\left( {4 - x} \right)\left( {x + 3} \right)\left( {1 + \sqrt {4 + x} } \right)}}{{ - \left( {\sqrt {x + 12}  - x} \right)\left( {x + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 3} \frac{{\left( {4 - x} \right)\left( {1 + \sqrt {4 + x} } \right)}}{{ - \left( {\sqrt {x + 12}  - x} \right)}}\\
 = \frac{{7.2}}{{ - \left( {3 + 3} \right)}} =  - \frac{{14}}{6}\\
f.\mathop {\lim }\limits_{x \to 1} \frac{{\left( {3x + 1 - 4{x^2}} \right)\left( {4 + 2\sqrt[3]{{\left( {8x} \right)}} + \sqrt[3]{{{{\left( {8x} \right)}^2}}}} \right)}}{{\left( {8 - 8x} \right)\left( {\sqrt {3x + 1}  + 2x} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {4x + 1} \right)\left( {1 - x} \right)\left( {4 + 2\sqrt[3]{{\left( {8x} \right)}} + \sqrt[3]{{{{\left( {8x} \right)}^2}}}} \right)}}{{8\left( {1 - x} \right)\left( {\sqrt {3x + 1}  + 2x} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {4x + 1} \right)\left( {4 + 2\sqrt[3]{{\left( {8x} \right)}} + \sqrt[3]{{{{\left( {8x} \right)}^2}}}} \right)}}{{8\left( {\sqrt {3x + 1}  + 2x} \right)}}\\
 = \frac{{5\left( {4 + 4 + 4} \right)}}{{8\left( {2 + 2} \right)}} = \frac{{15}}{8}
\end{array}\)