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

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(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\}

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Continuous everywhere

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(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).

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