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

We are asked to evaluate

$$\lim_{x \to 0^{+}} \; x\Bigl(\,[\tfrac1x]+[\tfrac2x]+\dots +[\tfrac{15}{x}]\,\Bigr),$$

where $$[\,t\,]$$ denotes the greatest integer less than or equal to $$t$$ (the floor function).

First, recall (state) the basic relation between any real number $$y$$, its greatest-integer part $$[y]$$ and its fractional part $$\{y\}$$:

$$y \;=\; [y] \;+\; \{y\}, \qquad\text{with } 0\le \{y\}<1.$$

Re-arranging this equality gives the useful identity

$$[y] \;=\; y \;-\; \{y\}.$$

We shall apply this identity to every term $$\bigl[\tfrac{k}{x}\bigr]$$ inside the bracketed sum, where $$k=1,2,\dots,15$$ and $$x>0$$ (because the limit approaches $$0$$ from the right).

For a fixed $$k$$ we therefore write

$$\Bigl[\tfrac{k}{x}\Bigr] \;=\; \frac{k}{x}\;-\;\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Multiplying both sides by $$x$$ (which is positive) gives

$$x\Bigl[\tfrac{k}{x}\Bigr] \;=\; x\!\left(\frac{k}{x}-\Bigl\{\tfrac{k}{x}\Bigr\}\right) = k - x\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Now we add these equalities for all $$k=1$$ to $$15$$:

$$x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] =\sum_{k=1}^{15}\!\Bigl(k - x\bigl\{\tfrac{k}{x}\bigr\}\Bigr) =\sum_{k=1}^{15}k \;-\; x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}.$$

The first summation is a simple arithmetic series. Using the standard formula

$$1+2+\dots+n=\frac{n(n+1)}{2},$$

with $$n=15$$, we have

$$\sum_{k=1}^{15} k \;=\;\frac{15\cdot16}{2}=120.$$

Hence our overall expression becomes

$$x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] = 120 \;-\; x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}.$$

Observe next that each fractional part satisfies $$0\le\{\,\cdot\,\}<1$$, so

$$0\;\le\;\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}\;\le\;15.$$ Multiplying by the positive number $$x$$ gives

$$0\;\le\;x\sum_{k=1}^{15}\Bigl\{\tfrac{k}{x}\Bigr\}\;\le\;15x.$$

As $$x\to 0^{+}$$ the quantity $$15x$$ tends to $$0$$, and by squeeze (sandwich) reasoning the product $$x\sum_{k=1}^{15}\bigl\{\tfrac{k}{x}\bigr\}$$ also tends to $$0$$.

Therefore, taking the limit of both sides, we get

$$\lim_{x\to0^{+}} x\sum_{k=1}^{15}\Bigl[\tfrac{k}{x}\Bigr] = 120 - 0 = 120.$$

Hence, the correct answer is Option D.