Đáp án:
16) -1
Giải thích các bước giải:
\(\begin{array}{l}
11)\lim \dfrac{{{{\left( {2 - \dfrac{1}{n}} \right)}^3}{{\left( {1 - \dfrac{3}{n}} \right)}^3}}}{{3{{\left( {1 + \dfrac{1}{n}} \right)}^9}}}\\
= \lim \dfrac{{{2^3}{{.1}^3}}}{{{{3.1}^9}}} = \dfrac{8}{3}\\
13)\lim \dfrac{{1 - \dfrac{2}{{{n^4}}} + \dfrac{1}{{{n^5}}}}}{{2 - \dfrac{1}{{{n^4}}} + \dfrac{3}{{{n^5}}}}} = \dfrac{1}{2}\\
15)\lim \dfrac{{\sqrt {3 + \dfrac{1}{{{n^2}}}} - \sqrt {1 - \dfrac{1}{{{n^2}}}} }}{1}\\
= \lim \dfrac{{\sqrt 3 - 1}}{1} = \sqrt 3 - 1\\
12)\lim \dfrac{{6 - \dfrac{2}{{{n^2}}} + \dfrac{1}{{{n^3}}}}}{{2 - \dfrac{1}{{{n^2}}}}} = \lim \dfrac{6}{2} = 3\\
14)\lim \dfrac{{\left( {4{n^2} + 3 - 4{n^2} + 4n - 1} \right)\left( {\sqrt {{n^2} + 2n} + n} \right)}}{{\left( {{n^2} + 2n - {n^2}} \right)\left( {\sqrt {4{n^2} + 3} + 2n - 1} \right)}}\\
= \lim \dfrac{{\left( {4n + 2} \right)\left( {\sqrt {{n^2} + 2n} + n} \right)}}{{2n\left( {\sqrt {4{n^2} + 3} + 2n - 1} \right)}}\\
= \lim \dfrac{{\left( {4 + \dfrac{2}{n}} \right)\left( {\sqrt {1 + \dfrac{2}{n}} + 1} \right)}}{{2\left( {\sqrt {4 + \dfrac{3}{{{n^2}}}} + 2 - \dfrac{1}{n}} \right)}}\\
= \lim \dfrac{{4.2}}{{2.\left( {2 + 2} \right)}} = 1\\
16)\lim \dfrac{{\sqrt {1 + \dfrac{1}{n} - \dfrac{1}{{{n^2}}}} - \sqrt {4 - \dfrac{2}{{{n^2}}}} }}{{1 + \dfrac{3}{n}}}\\
= \dfrac{{1 - 2}}{1} = - 1
\end{array}\)
( bạn xem lại đề câu 9 và 10 nhé )
