If [x] denotes the greatest integer leq x,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$. Substituting this into the limit,we get: $\lim_{n

Vedclass · Accepted Answer

$\frac{x}{3}$

Vedclass · Answer

(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$. Substituting this into the limit,we get: $\lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n [r^2 x] = \lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n (r^2 x - \{r^2 x\})$ $= \lim_{n \rightarrow \infty} \left( \frac{x}{n^3} \sum_{r=1}^n r^2 - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$ Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$,we have: $= \lim_{n \rightarrow \infty} \left( \frac{x \cdot n(n+1)(2n+1)}{6n^3} - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$ Since $0 \leq \{r^2 x\} Thus,the limit is $\lim_{n \rightarrow \infty} \frac{x(2n^3 + 3n^2 + n)}{6n^3} = \frac{2x}{6} = \frac{x}{3}$.

If $[x]$ denotes the greatest integer $\leq x$,then $\lim_{n \rightarrow \infty} \frac{1}{n^3} \{[1^2 x] + [2^2 x] + [3^2 x] + \ldots + [n^2 x] \} = $

Similar Questions

Evaluate the limit: $\mathop {\lim }\limits_{x \to 3} (x+3)$

Let $m$ and $n$ be two positive integers greater than $1$. If $\lim_{\alpha \rightarrow 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\left( \frac{e}{2} \right)$,then the value of $\frac{m}{n}$ is

$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}} = $

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 1} \frac{4x+3}{x-2}$

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module