AP EAMCET · Maths · Limits
If \([x]\) denotes the greatest integer \(\leq x\), then \(\lim _{n \rightarrow-\infty} \frac{1}{n^3}\left\{\left[1^2 x\right]+\left[2^2 x\right]+\left[3^2 x\right]+\ldots+\left[n^2 x\right]\right\}=\)
- A \(\frac{x}{2}\)
- B \(\frac{x}{3}\)
- C \(\frac{x}{6}\)
- D 0
Answer & Solution
Correct Answer
(B) \(\frac{x}{3}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \lim _{n \rightarrow \infty} \frac{1}{n^3}\left\{\left[1^2 x\right]+\left[2^2 x\right]+\left[3^2 x\right]+\ldots+\left[n^2 x\right]\right\} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n\left[r^2 x\right]=\lim _{n \rightarrow \infty}…
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