Đáp án:
1. lim =0
2. lim=0
3. lim=1/3
4. lim=0
5. lim=1
Giải thích các bước giải:
\(\begin{array}{l}
1)\lim \frac{{\sqrt[3]{{\frac{1}{{{n^2}}}}} + \frac{2}{n}}}{{1 + \frac{1}{{\sqrt n }}}} = \frac{0}{1} = 0\\
2)\lim \frac{{\sqrt[3]{{\frac{1}{n} + \frac{3}{{{n^2}}} + \frac{2}{{{n^3}}}}}}}{{\sqrt {1 - \frac{4}{n} + \frac{5}{{{n^2}}}} }} = \frac{0}{1} = 0\\
3)\lim \frac{{\sqrt {{n^4} + {n^3}} }}{{3{n^2} + n - 2}}\\
= \lim \frac{{\sqrt {1 + \frac{1}{n}} }}{{3 + \frac{1}{n} - \frac{2}{{{n^2}}}}} = \frac{1}{3}\\
4)\lim \frac{{\sqrt[3]{{\frac{1}{{{n^3}}} + \frac{1}{{{n^6}}}}} + \frac{1}{{\sqrt n }}}}{{1.\sqrt {1 + \frac{1}{{{n^2}}}} }} = \frac{0}{1} = 0\\
5)\lim \frac{{1 + \sqrt[3]{{\frac{1}{{{n^3}}} + \frac{1}{{{n^6}}}}} + \frac{1}{{\sqrt n }}}}{{1.\sqrt {1 + \frac{1}{{{n^2}}}} + \frac{3}{{{n^2}}}}} = \frac{1}{1} = 1
\end{array}\)
