Question

For each $t\in R,$ let [t] be the greatest integer less than or equal to $t$. then $\underset{x\to 0^{+}}{lim}x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+........+\left[\frac{15}{x}\right]\right)$

• A. does not exist (in R)
• B. is equal to $0$
• C. is equal to $15$
• D. is equal to $120$

Answer 1

The correct option is D is equal to $120$

$\begin{array}{c}Given,\\ {lim}_{x\to 0}x\left[\left(\frac{1}{x}\right)+\left(\frac{2}{x}\right)+..........+\left(\frac{15}{x}\right)\right]\\ Since\left[x\right]=x-\left\{x\right\}\\ {lim}_{x\to 0}x\left[\frac{1}{x}+\frac{2}{x}+..........+\frac{15}{x}=\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+..........\left\{\frac{15}{x}\right\}\right]\\ {lim}_{x\to 0}x\left(1+2+.......+15\right)-\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}++..........\left\{\frac{15}{x}\right\}\\ =\frac{15\left(15+1\right)}{2}-0\\ =15\times 8=120\end{array}$

Hence, this is the answer.