Đáp án:
\(\lim \frac{{{{\left( {n + 3} \right)}^{40}}.{{\left( {4{n^3} - 1} \right)}^{20}}}}{{{{\left( {3 + 2n} \right)}^{50}}.{{\left( {1 + {n^2}} \right)}^{25}}}} = \frac{1}{{{2^{10}}}}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{{{\left( {n + 3} \right)}^{40}}.{{\left( {4{n^3} - 1} \right)}^{20}}}}{{{{\left( {3 + 2n} \right)}^{50}}.{{\left( {1 + {n^2}} \right)}^{25}}}}\\
= \lim \frac{{\frac{{{{\left( {n + 3} \right)}^{40}}}}{{{n^{40}}}}.\frac{{{{\left( {4{n^3} - 1} \right)}^{20}}}}{{{n^{60}}}}}}{{\frac{{{{\left( {3 + 2n} \right)}^{50}}}}{{{n^{50}}}}.\frac{{{{\left( {1 + {n^2}} \right)}^{25}}}}{{{n^{50}}}}}}\\
= \lim \frac{{{{\left( {1 + \frac{3}{n}} \right)}^{40}}.{{\left( {4 - \frac{1}{{{n^3}}}} \right)}^{20}}}}{{{{\left( {\frac{3}{n} + 2} \right)}^{50}}.{{\left( {\frac{1}{{{n^2}}} + 1} \right)}^{25}}}}\\
= \frac{{{1^{40}}{{.4}^{20}}}}{{{2^{50}}{{.1}^{25}}}} = \frac{{{2^{40}}}}{{{2^{50}}}} = \frac{1}{{{2^{10}}}}
\end{array}\)
