Đáp án:
f) \(\dfrac{1}{{48}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\lim \dfrac{{1 + \dfrac{2}{{{3^n}}}}}{{5 + {{\left( {\dfrac{2}{3}} \right)}^n}}} = \dfrac{1}{5}\\
b)\lim \dfrac{{\left( {3 + \dfrac{1}{n}} \right)\left( {\dfrac{3}{n} - 7} \right)}}{{{{\left( {1 + \dfrac{1}{n}} \right)}^2}}} = \dfrac{{3.\left( { - 7} \right)}}{1} = - 21\\
c)\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 6} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 6}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{1 + 6}}{{1 + 1}} = \dfrac{7}{2}\\
d)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{4 - \sqrt {1 + \dfrac{1}{{{n^2}}}} }}{{7 + \dfrac{1}{n}}} = \dfrac{{4 - 1}}{7} = \dfrac{3}{7}\\
f)\mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 6 - 4}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {\sqrt {x + 6} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{{x + 2}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {\sqrt {x + 6} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{1}{{\left( {{x^2} - 2x + 4} \right)\left( {\sqrt {x + 6} + 2} \right)}}\\
= \dfrac{1}{{\left( {{{\left( { - 2} \right)}^2} - 2.\left( { - 2} \right) + 4} \right)\left( {\sqrt { - 2 + 6} + 2} \right)}} = \dfrac{1}{{48}}
\end{array}\)
