If f(x) = lim_{n to infty} frac{[x^2] + [(2x) | vedclass

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

Question

(A) We are given $f(x) = \lim_{n 	o \infty} \frac{\sum_{k=1}^{n} [(kx)^2]}{n^3}$.Using the property of the Greatest Integer Function,$[y] = y - \{y\}

Vedclass · Accepted Answer

Continuous everywhere

Vedclass · Answer

(A) We are given $f(x) = \lim_{n 	o \infty} \frac{\sum_{k=1}^{n} [(kx)^2]}{n^3}$.Using the property of the Greatest Integer Function,$[y] = y - \{y\}$,where $0 \le \{y\} $f(x) = \lim_{n 	o \infty} \frac{\sum_{k=1}^{n} ((kx)^2 - \{(kx)^2\})}{n^3}$.$f(x) = \lim_{n 	o \infty} \left( \frac{x^2 \sum_{k=1}^{n} k^2}{n^3} - \frac{\sum_{k=1}^{n} \{(kx)^2\}}{n^3} ight)$.Since $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,the first term becomes $\lim_{n 	o \infty} \frac{x^2 n(n+1)(2n+1)}{6n^3} = \frac{x^2}{6} 	imes 2 = \frac{x^2}{3}$.The second term involves $\sum_{k=1}^{n} \{(kx)^2\}$. Since $0 \le \{(kx)^2\} Therefore,$f(x) = \frac{x^2}{3}$.Since $f(x) = \frac{x^2}{3}$ is a polynomial function,it is continuous for all $x \in R$.

If $f(x) = \lim_{n \to \infty} \frac{[x^2] + [(2x)^2] + [(3x)^2] + \cdots + [(nx)^2]}{n^3}$,then $f(x)$ is (Where $[\cdot]$ is the Greatest Integer Function).

Similar Questions

Examine whether the function $f$ given by $f(x) = x^{2}$ is continuous at $x = 0$.

If the function $f(x) = \begin{cases} (\cos x)^{1/x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

The function $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \right) \pi$,where $[\cdot]$ denotes the greatest integer function,is discontinuous at

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

Let $f(x) = \begin{cases} |x|+3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x+2, & \text{if } x \geq 3 \end{cases}$. Determine the continuity of $f(x)$ at $x = -3$ and $x = 3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module