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