Giải thích các bước giải:
\(\begin{array}{l}
b,\\
\lim \frac{{\left( {2n\sqrt n + 1} \right)\left( {\sqrt n + 3} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\\
= \lim \dfrac{{\frac{{\left( {2n\sqrt n + 1} \right)\left( {\sqrt n + 3} \right)}}{{{n^2}}}}}{{\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{{{n^2}}}}}\\
= \lim \dfrac{{\frac{{2n\sqrt n + 1}}{{n\sqrt n }}.\frac{{\sqrt n + 3}}{{\sqrt n }}}}{{\frac{{n + 1}}{n}.\frac{{n + 2}}{n}}}\\
= \lim \dfrac{{\left( {2 + \frac{1}{{n\sqrt n }}} \right)\left( {1 + \frac{3}{{\sqrt n }}} \right)}}{{\left( {1 + \frac{1}{n}} \right)\left( {1 + \frac{2}{n}} \right)}}\\
= \frac{{2.1}}{{1.1}} = 2\\
c,\\
\lim \frac{{{n^2} + \sqrt[3]{{1 - {n^6}}}}}{{\sqrt {{n^4} + 1} - {n^2}}}\\
= \lim \left[ {\frac{{{{\left( {{n^2}} \right)}^3} + \left( {1 - {n^6}} \right)}}{{{{\left( {{n^2}} \right)}^2} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}:\frac{{\left( {{n^4} + 1} \right) - {{\left( {{n^2}} \right)}^2}}}{{\sqrt {{n^4} + 1} + {n^2}}}} \right]\\
= \lim \left[ {\frac{1}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}:\frac{1}{{\sqrt {{n^4} + 1} + {n^2}}}} \right]\\
= \lim \frac{{\sqrt {{n^4} + 1} + {n^2}}}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}\\
= \lim \frac{{\sqrt {1 + \frac{1}{{{n^4}}}} + 1}}{{{n^2} - \sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{\frac{1}{{{n^3}}} - {n^3}}}}^2}}}\\
= \frac{2}{{ + \infty }} = 0\\
d,\\
\lim \frac{{\sqrt[3]{{8{n^3} + 1}}}}{{2n - 5}}\\
= \lim \frac{{\sqrt[3]{{8 + \frac{1}{{{n^3}}}}}}}{{2 - \frac{5}{n}}}\\
= \frac{{\sqrt[3]{8}}}{2} = 1
\end{array}\)
