Question

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{\left[x^2\right]+\left[{\left(2x\right)}^2\right]+..+\left[{\left(nx\right)}^2\right]}{n^3}$ then the set of all points of continuity of $f$ (where $\left[x\right]$ denotes the greatest integer function)

• A. $(-\infty ,\infty )-\left\{0\right\}$
• B. $(-\infty ,\infty )-I$
• C. $(-\infty ,\infty )$
• D. $(-\infty ,\infty )-\{0,1\}$

Answer 1

The correct option is C $(-\infty ,\infty )$

By the definition of the greatest integer function, $\left[y\right]=y-k$ where $0\le k<1$. Using this,
$f\left(x\right)=\underset{n\to \infty }{lim}\frac{\left[x^2\right]+\left[{\left(2x\right)}^2\right]+\cdots +\left[{\left(nx\right)}^2\right]}{n^3}$
$=\underset{n\to \infty }{lim}\frac{x^2-k_1+{\left(2x\right)}^2-k_2+{\left(3x\right)}^2-k_3+\cdots +{\left(nx\right)}^2-k_n}{n^3}$
$=\underset{n\to \infty }{lim}(\frac{x^2+{\left(2x\right)}^2+{\left(3x\right)}^2+\cdots +{\left(nx\right)}^2}{n^3}-\frac{k_1+k_2+k_3+\cdots +k_n}{n^3})$
$=\underset{n\to \infty }{lim}\frac{x^2+{\left(2x\right)}^2+{\left(3x\right)}^2+\cdots +{\left(nx\right)}^2}{n^3}-(k_1+k_2+k_3+\cdots +k_n)\underset{n\to \infty }{lim}\frac{1}{n^3}$
$=\underset{n\to \infty }{lim}\frac{x^2(1^2+2^2+\cdots +n^2)}{n^3}$
The second term in this expression is zero. So, the limit becomes
$f\left(x\right)=\underset{n\to \infty }{lim}\frac{x^2(1^2+2^2+\cdots +n^2)}{n^3}=x^2\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6n^3}=x^2\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}=\frac{x^2}{3}$
This function is continuous everywhere. So the answer is option C.