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

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

Let $f(x) = [x]\sin \left( \frac{\pi}{[x + 1]} \right)$,where $[.]$ denotes the greatest integer function. The domain of $f$ is and the points of discontinuity of $f$ in the domain are

Let $f(x) = x^2 + x \sin x - \cos x$. Then

Consider the function $f:(0,2) \rightarrow R$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\begin{cases} \min \{f(t) : 0 < t \leq x\}, & 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{cases}$. Then,

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module