Given,

$$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $$ $$\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$$

as we know that

$${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$$

$$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$$

We know $$\left\{ {{1 \over x}} \right\}$$ is fractional part of $${1 \over x}.$$

So, the range of $$\,\left\{ {{1 \over x}} \right\}$$ is $$0 \le \left\{ {{1 \over x}} \right\} < 1$$

So, $$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$$ (finite no) $$=0$$

Similarly $$\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$$

$$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$$

$$ = \,\,\,\,{{15 \times 16} \over 2}$$

$$ = \,\,\,\,120$$