Question

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

Answer 1

Finding the value of $\underset{\mathbf{x}\mathbf{\to }{\mathbf{0}}^{\mathbf{+}}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{x}\mathbf{(}\mathbf{}\mathbf{[}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\mathbf{}\mathbf{+}\mathbf{[}\mathbf{}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{]}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\frac{\mathbf{15}}{\mathbf{x}}\mathbf{\left]}\mathbf{\right)}$

Given that $t\in R,$

$\begin{array}{rcl}& & \underset{x\to 0^{+}}{\lim }x([\frac{1}{x}]-[\frac{1}{x}]+[\frac{2}{x}]-[\frac{2}{x}]+....[\frac{15}{x}]-[\frac{15}{x}\left]\right)\\ & =& \underset{x\to 0^{+}}{\lim }(1+2+3+....+15-x(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+....+\left[\frac{15}{x}\right]\left)\right)\\ & =& 120-\underset{x\to 0^{+}}{\lim }x([\frac{1}{x}]+[\frac{2}{x}]+...+[\frac{15}{x}])\{finitevalue\}\\ & =& 120\end{array}$

Hence the value of $\underset{\mathbf{x}\mathbf{\to }{\mathbf{0}}^{\mathbf{+}}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{x}\mathbf{(}\mathbf{}\mathbf{[}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\mathbf{}\mathbf{+}\mathbf{[}\mathbf{}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{]}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\frac{\mathbf{15}}{\mathbf{x}}\mathbf{\left]}\mathbf{\right)}$ is $\mathbf{120}\mathbf{.}$