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(B) We know that $[t] = t - \{t\}$,where $\{t\}$ is the fractional part of $t$.Thus,the expression becomes $\lim_{x 	o 0^+} x \sum_{r=1}^{15} [\frac{

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$120$

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(B) We know that $[t] = t - \{t\}$,where $\{t\}$ is the fractional part of $t$.Thus,the expression becomes $\lim_{x 	o 0^+} x \sum_{r=1}^{15} [\frac{r}{x}] = \lim_{x 	o 0^+} x \sum_{r=1}^{15} (\frac{r}{x} - \{\frac{r}{x}\})$.$= \lim_{x 	o 0^+} (\sum_{r=1}^{15} r - x \sum_{r=1}^{15} \{\frac{r}{x}\})$.Since $0 \le \{\frac{r}{x}\} As $x 	o 0^+$,by the Squeeze Theorem,$x \{\frac{r}{x}\} 	o 0$.Therefore,the limit is $\sum_{r=1}^{15} r = \frac{15 	imes 16}{2} = 120$.

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

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