Đáp án:
e) \(1 + \sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
d)\lim \left( {\dfrac{{{n^3} + 1}}{n}} \right)\left( {\dfrac{{3 - 2n}}{{{n^3} - 2}}} \right)\\
= \lim \left( {\dfrac{{1 + \dfrac{1}{{{n^3}}}}}{1}} \right)\left( {\dfrac{{\dfrac{3}{n} - 2}}{{1 - \dfrac{2}{{{n^3}}}}}} \right)\\
= \lim \dfrac{{1.\left( { - 2} \right)}}{{1.1}} = - 2\\
e)\lim \dfrac{{\sqrt {1 - \dfrac{2}{n}} + \sqrt {3 + \dfrac{1}{{{n^2}}}} }}{{1 + \dfrac{3}{n}}}\\
= \dfrac{{1 + \sqrt 3 }}{1} = 1 + \sqrt 3
\end{array}\)
