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$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)=$
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The correct answer is:
$\frac{x}{3}$
$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n^3}\left(1^2+2^2+\ldots .+n^2\right) x \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1)}{6} \times x \\ & \lim _{n \rightarrow \infty} \frac{x}{6}\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2 x}{6}=\frac{x}{3}\end{aligned}$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n^3}\left(1^2+2^2+\ldots .+n^2\right) x \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \frac{n(n+1)(2 n+1)}{6} \times x \\ & \lim _{n \rightarrow \infty} \frac{x}{6}\left(2+\frac{3}{n}+\frac{1}{n^2}\right)=\frac{2 x}{6}=\frac{x}{3}\end{aligned}$
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