Đáp án:
\[0\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
A = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{{n^2}}}{{{n^3} + 1}} + \frac{{{n^2}}}{{{n^3} + 2}} + \frac{{{n^2}}}{{{n^3} + 3}} + ..... + \frac{{{n^2}}}{{{n^3} + n}}} \right)\\
= \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{{{n^2}}}}} + \frac{1}{{n + \frac{2}{{{n^2}}}}} + \frac{1}{{n + \frac{3}{{{n^2}}}}} + ..... + \frac{1}{{n + \frac{1}{n}}}} \right)\\
\mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{{{n^2}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{2}{{{n^2}}}} \right) = ..... = \mathop {\lim }\limits_{n \to + \infty } \left( {n + \frac{1}{n}} \right) = + \infty \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{1}{{{n^2}}}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{1 + \frac{2}{{{n^2}}}}}} \right) = .... = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{1}{{n + \frac{1}{n}}}} \right) = 0\\
\Rightarrow A = 0
\end{array}\]
