Question

Evaluate $\underset{n\to \infty }{lim}\frac{[1.x]+[2.x]+[3.x]+.......+[n.x]}{n^2}$, where $[.]$ denotes the greatest integer function.

• A. $\frac{-x}{2}$
• B. $\frac{x}{2}$
• C. $\frac{x}{4}$
• D. $\frac{-x}{3}$

Answer 1

Answer: (B)

$\frac{x}{2}$

Let $f\left(x\right)=\frac{\left[x\right]+\left[2x\right]+\left[3x\right]+...+\left[nx\right]}{n^2}$
Now we have
$f\left(x\right)\le \frac{x+2x+3x+...+nx}{n^2}=\frac{x\left(n\right)(n+1)}{{2n}^2}=\frac{x}{2}(1+\frac{1}{n})$
And $f\left(x\right)>\frac{(x-1)+(2x-1)...(nx-1)}{n^2}=\frac{\frac{xn(n+1)}{2}-n}{n^2}=\frac{x}{2}(1+\frac{1}{n})-\frac{1}{n}$
Therefore from sandwich theorem
$\frac{x}{2}(1+\frac{1}{n})-\frac{1}{n}<f\left(x\right)\le \frac{x}{2}(1+\frac{1}{n})$
$\Rightarrow \underset{n\to \infty }{lim}\left(\frac{x}{2}(1+\frac{1}{n})\frac{-1}{n}\right)<\underset{x\to \infty }{lim}f\left(x\right)\le \underset{x\to \infty }{lim}\frac{x}{2}(1+\frac{1}{n})$
$\Rightarrow \frac{x}{2}<\underset{n\to \infty }{lim}f\left(x\right)\le \frac{x}{2}$
$\therefore \underset{n\to \infty }{lim}f\left(x\right)=\frac{x}{2}$