We need to evaluate $L=\lim_{x\to 0^+}x\left(\left[\frac1x\right]+\left[\frac2x\right]+\cdots+\left[\frac{15}{x}\right]\right).$

Use the basic property of greatest integer function: $y-1<[y]\le y.$ Applying this to each term with $y=\dfrac{k}{x}$, for $k=1,2,\dots,15$, we get $\frac{k}{x}-1<\left[\frac{k}{x}\right]\le \frac{k}{x}.$

Multiply throughout by $x>0$: $k-x<x\left[\frac{k}{x}\right]\le k.$ This is true for each $k=1,2,\dots,15$.

Sum these inequalities from $k=1$ to $15$: $\sum_{k=1}^{15}(k-x)<x\sum_{k=1}^{15}\left[\frac{k}{x}\right]\le \sum_{k=1}^{15}k.$ That is, $\left(\sum_{k=1}^{15}k\right)-15x<x\left(\left[\frac1x\right]+\left[\frac2x\right]+\cdots+\left[\frac{15}{x}\right]\right)\le \sum_{k=1}^{15}k.$

Compute the sum: $\sum_{k=1}^{15}k=\frac{15\cdot 16}{2}=120.$ So, $120-15x<x\left(\left[\frac1x\right]+\left[\frac2x\right]+\cdots+\left[\frac{15}{x}\right]\right)\le 120.$

Now let $x\to 0^+$. Then $120-15x\to 120,$ and the upper bound is already $120$. Hence, by the squeeze theorem, $L=120.$

Therefore the correct option is $\boxed{\text{D}}.$