For each$t\in R$, let  $\left[t\right]$be the greatest integer less than or equal to ‘t’. Then $\underset{x\to 0+}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{........}+\left[\frac{10}{x}\right]\right)$

$\begin{array}{l}\frac{1}{x}-1<\left[\frac{1}{x}\right]\le \frac{1}{x}\\ \frac{2}{x}-1<\left[\frac{2}{x}\right]\le \frac{2}{x}\\ \frac{10}{x}-1<\left[\frac{10}{x}\right]\le \frac{10}{x}\end{array}$

$\begin{array}{c}\frac{(1-x)+(2-x)+\mathrm{.....}+(10-x)}{x}<\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{......}+\left[\frac{10}{x}\right]\le \frac{1+2+\mathrm{.....}+10}{x}\\ \frac{55-10x}{x}<\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{......}+\left[\frac{10}{x}\right]\le \frac{55}{x}\\ \underset{x\to 0+}{\lim }55-10x<\underset{x\to 0+}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{......}+\left[\frac{10}{x}\right]\right)\le \underset{x\to 0+}{\lim }55\\ 55<\underset{x\to 0+}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{......}+\left[\frac{10}{x}\right]\right)\le 55\\ \underset{x\to 0+}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{........}+\left[\frac{10}{x}\right]\right)\\ =55\end{array}$