lim_n –>vô cực (n^2/n^3+1+n^2/n^3+2+…+n^2/n^3+n) 26/07/2021 Bởi Gabriella lim_n –>vô cực (n^2/n^3+1+n^2/n^3+2+…+n^2/n^3+n)
Đáp án: \[0\] Giải thích các bước giải: Ta có: \[\begin{array}{l}A = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{{n^2}}}{{{n^3} + 1}} + \frac{{{n^2}}}{{{n^3} + 2}} + \frac{{{n^2}}}{{{n^3} + 3}} + ….. + \frac{{{n^2}}}{{{n^3} + n}}} \right)\\ = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{{{n^2}}}}} + \frac{1}{{n + \frac{2}{{{n^2}}}}} + \frac{1}{{n + \frac{3}{{{n^2}}}}} + ….. + \frac{1}{{n + \frac{1}{n}}}} \right)\\\mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{{{n^2}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{2}{{{n^2}}}} \right) = ….. = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{n}} \right) = + \infty \\ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{1}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{2}{{{n^2}}}}}} \right) = …. = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{n}}}} \right) = 0\\ \Rightarrow A = 0\end{array}\] Bình luận
Đáp án:
\[0\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
A = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{{n^2}}}{{{n^3} + 1}} + \frac{{{n^2}}}{{{n^3} + 2}} + \frac{{{n^2}}}{{{n^3} + 3}} + ….. + \frac{{{n^2}}}{{{n^3} + n}}} \right)\\
= \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{{{n^2}}}}} + \frac{1}{{n + \frac{2}{{{n^2}}}}} + \frac{1}{{n + \frac{3}{{{n^2}}}}} + ….. + \frac{1}{{n + \frac{1}{n}}}} \right)\\
\mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{{{n^2}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{2}{{{n^2}}}} \right) = ….. = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{n}} \right) = + \infty \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{1}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{2}{{{n^2}}}}}} \right) = …. = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{n}}}} \right) = 0\\
\Rightarrow A = 0
\end{array}\]