Đáp án:
d) 2
Giải thích các bước giải:
\(\begin{array}{l}
a)\lim n.\dfrac{{ - 6n - 2 + \dfrac{7}{n}}}{{ - 2n + 3}}\\
= \lim n.\dfrac{{ - 6 - \dfrac{2}{n} + \dfrac{7}{{{n^2}}}}}{{ - 2 + \dfrac{3}{n}}} = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{ - 6 - \dfrac{2}{n} + \dfrac{7}{{{n^2}}}}}{{ - 2 + \dfrac{3}{{{n^2}}}}} = 3\\
b)\lim \dfrac{{3 + \dfrac{2}{n} + \dfrac{4}{{{n^2}}}}}{{2.\left( {1 + \dfrac{1}{n}} \right)}} = \dfrac{3}{2}\\
c)\lim \dfrac{{3 + \dfrac{2}{n}.\left( { - 1 + \dfrac{5}{{{n^2}}}} \right)}}{{1 + \dfrac{3}{{{n^2}}} - \dfrac{{11}}{{{n^3}}}}} = 3\\
d)\lim \dfrac{{{{\left( {2 + \dfrac{3}{n}} \right)}^2}}}{{2 + \dfrac{3}{{{n^2}}}}} = \dfrac{4}{2} = 2
\end{array}\)
