Q102 / 160 · TS EAMCET 2004 · Mathematics

The value of $\lim _{\substack{n \rightarrow \infty}} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)$ is

  • A
    $x$
  • B
    $\frac{x}{2}$
  • Correct answer
    C
    $\frac{x}{3}$
  • D
    $\frac{x}{4}$

Solution

$\begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2\right. & x) \\ & =x \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n\left(\frac{k}{n}\right)^2 \\ & =x \int_0^1 x^2 d x=x\left[\frac{x^3}{3}\right]_0^1 \\ & =\frac{x}{3}\end{aligned}$

Hence, the correct answer is option: 'C'.

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