Mathematics - Limits Question with Solution | TestHub

We know that

$\frac{1}{x}-1<\left[\frac{1}{x}\right]\le \frac{1}{x}$

Hence, 

$\frac{1+2+3.\; .\; .+15}{x}-15<\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\mathrm{....}+\left[\frac{15}{x}\right]\le \frac{1+2+\mathrm{...}+15}{x}$

$\Rightarrow \underset{x\to 0^{+}}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+....+\left[\frac{15}{x}\right]\right)=15\left(\frac{16}{2}\right)$

$=120$