Đáp án:

g) 6

Giải thích các bước giải:

\(\begin{array}{l}
b)\lim \dfrac{{{{2.4.4}^n} - {{3.3}^n}}}{{4 + {5^{ - 2}}{{.5}^n}}} = \lim \dfrac{{8.{{\left( {\dfrac{4}{5}} \right)}^n} - 3.{{\left( {\dfrac{3}{5}} \right)}^n}}}{{\dfrac{4}{{{5^n}}} + {5^{ - 2}}}}\\
 = \dfrac{0}{{{5^{ - 2}}}} = 0\\
e)\mathop {\lim }\limits_{x \to  - \dfrac{1}{2}} \dfrac{{2{x^2} - x - 1}}{{1 + 2x}} = \mathop {\lim }\limits_{x \to  - \dfrac{1}{2}} \dfrac{{\left( {2x + 1} \right)\left( {x - 1} \right)}}{{2x + 1}}\\
 = \mathop {\lim }\limits_{x \to  - \dfrac{1}{2}} \left( {x - 1} \right) =  - \dfrac{3}{2}\\
f)\mathop {\lim }\limits_{x \to 1} \dfrac{{2x + 7 - {{\left( {4x - 1} \right)}^2}}}{{\left( {1 - x} \right)\left( {\sqrt {2x + 1}  + 4x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 16{x^2} + 8x - 1 + 2x + 7}}{{\left( {1 - x} \right)\left( {\sqrt {2x + 1}  + 4x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {1 - x} \right)\left( {8x + 3} \right)}}{{\left( {1 - x} \right)\left( {\sqrt {2x + 1}  + 4x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{8x + 3}}{{\sqrt {2x + 1}  + 4x - 1}} = \dfrac{{33 - 11\sqrt 3 }}{6}\\
g)\mathop {\lim }\limits_{x \to  - 2} \dfrac{{{{\left( {{x^3}} \right)}^2} - {{\left( {{2^3}} \right)}^2}}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\left( {{x^3} - {2^3}} \right)\left( {{x^3} + {2^3}} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \dfrac{{\left( {{x^2} + 2x + 4} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {{x^2} + 4} \right)}}\\
 = \dfrac{{\left( {{{\left( { - 2} \right)}^2} + 2.\left( { - 2} \right) + 4} \right)\left( {{{\left( { - 2} \right)}^2} - 2.\left( { - 2} \right) + 4} \right)}}{{\left( {{{\left( { - 2} \right)}^2} + 4} \right)}} = 6
\end{array}\)