Đáp án:
e) 1
Giải thích các bước giải:
\(\begin{array}{l}
a)\lim \dfrac{{5 - \dfrac{6}{n}}}{{3 + \dfrac{1}{n}}} = \dfrac{5}{3}\\
b)\lim \dfrac{{2 - \dfrac{1}{n} - \dfrac{3}{{{n^2}}}}}{{3 + \dfrac{2}{n} - \dfrac{1}{{{n^2}}}}} = \dfrac{2}{3}\\
c)\lim \dfrac{{\left( {2 - \dfrac{1}{n}} \right)\left( {1 + \dfrac{3}{n}} \right)}}{{1 - \dfrac{1}{n} - \dfrac{2}{{{n^2}}}}} = \dfrac{{2.1}}{1} = 2\\
d)\lim \dfrac{{\left( {3 - \dfrac{1}{n}} \right)\left( {1 + \dfrac{1}{n} + \dfrac{2}{{{n^2}}}} \right)}}{{2 - \dfrac{1}{{{n^2}}} - \dfrac{1}{{{n^3}}}}} = \dfrac{{3.1}}{2} = \dfrac{3}{2}\\
e)\lim \dfrac{{\sqrt {4 - \dfrac{1}{n} - \dfrac{6}{{{n^2}}}} }}{{2 + \dfrac{1}{n}}} = \dfrac{2}{2} = 1\\
f)\lim n\left( {\dfrac{{\sqrt {{n^2} - 2n - \dfrac{1}{n} - \dfrac{3}{{{n^2}}}} }}{{2n + 6}}} \right)\\
= \lim n\left( {\dfrac{{\sqrt {1 - \dfrac{2}{n} - \dfrac{1}{{{n^2}}} - \dfrac{3}{{{n^3}}}} }}{{2 + \dfrac{6}{n}}}} \right) = + \infty \\
Do:\mathop {\lim }\limits_{x \to + \infty } n = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {\dfrac{{\sqrt {1 - \dfrac{2}{n} - \dfrac{1}{{{n^2}}} - \dfrac{3}{{{n^3}}}} }}{{2 + \dfrac{6}{n}}}} \right) = \dfrac{1}{2}
\end{array}\)
