Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
\lim \frac{{\sqrt {{n^2} + 3} - n - 4}}{{\sqrt {{n^2} + 2} + n}}\\
= \lim \frac{{\frac{{{n^2} + 3 - {{\left( {n + 4} \right)}^2}}}{{\sqrt {{n^2} + 3} + n + 4}}}}{{\sqrt {{n^2} + 2} + n}}\\
= \lim \frac{{ - 8n - 13}}{{\left( {\sqrt {{n^2} + 3} + n + 4} \right).\left( {\sqrt {{n^2} + 2} + n} \right)}}\\
= \lim \frac{{ - \frac{8}{n} - \frac{{13}}{{{n^2}}}}}{{\left( {\sqrt {1 + \frac{3}{{{n^2}}}} + 1 + \frac{4}{n}} \right)\left( {\sqrt {1 + \frac{2}{{{n^2}}}} + 1} \right)}}\\
= \frac{0}{{2.2}} = 0\\
c,\\
\lim \frac{{{n^2} + \sqrt[3]{{1 - {n^6}}}}}{{\sqrt {{n^4} + 1} + {n^2}}}\\
= \lim \frac{{\frac{{{{\left( {{n^2}} \right)}^3} + {{\left( {\sqrt[3]{{1 - {n^6}}}} \right)}^3}}}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}}}{{\sqrt {{n^4} + 1} + {n^2}}}\\
= \lim \frac{1}{{\left( {{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}} \right).\left( {\sqrt {{n^4} + 1} + {n^2}} \right)}}\\
= 0
\end{array}\)
