Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
1.\lim \frac{{ - 13}}{{\sqrt n + \sqrt {13 + n} }} = \lim \frac{{\frac{{ - 13}}{{\sqrt n }}}}{{1 + \sqrt {\frac{{13}}{n} + 1} }} = - \infty \\
2.\lim \frac{{3n + 2}}{{\sqrt {9{n^2} + 3n + 2} + 3n}} = \lim \frac{{3 + \frac{2}{n}}}{{\sqrt {9 + \frac{3}{n} + \frac{2}{{{n^2}}}} + 3}} = \frac{3}{6} = \frac{1}{2}\\
3.\lim \frac{{4{n^2} + 32n + 64 - 4{n^2} - 12n - 5}}{{\left( {2n + 8} \right) + \sqrt {4{n^2} + 12n + 5} }} = \lim \frac{{20 + \frac{{59}}{n}}}{{2 + \frac{8}{n} + \sqrt {4 + \frac{{12}}{n} + \frac{5}{{{n^2}}}} }}\\
= \frac{{20}}{{2 + 2}} = 5\\
4.\lim \frac{{3{n^2} + {n^3} - {n^3}}}{{\left( {\sqrt[3]{{{{\left( {3{n^2} + {n^3}} \right)}^2}}} + n\sqrt[3]{{3{n^2} + {n^3}}} + {n^2}} \right)}}\\
= \lim \frac{3}{{\sqrt[3]{{\frac{{9{n^4} + 6{n^5} + {n^6}}}{{{n^6}}}}} + \frac{{n\sqrt[3]{{3{n^2} + {n^3}}}}}{{{n^2}}} + 1}} = \frac{3}{{1 + 1 + 1}} = 1
\end{array}\)
