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

Option: 1

120


Option: 2

115


Option: 3

132


Option: 4

112


Answers (1)

best_answer

Observe that \mathrm{t-1<[t] \leq t}
Applying this to numbers \mathrm{\frac{1}{x}, \frac{2}{x}, \ldots . . ., \frac{15}{x}} and summing them, we have

\mathrm{\sum_{k=1}^{15} \frac{k}{x}-15<\sum_{k=1}^{15}\left[\frac{k}{x}\right] \leq \sum_{k=1}^{15} \frac{k}{x}}

Multiplying throughout by \mathrm{x}, we have

\mathrm{ \sum_{k=1}^{15} k-15 x<x \sum_{k=1}^{15}\left[\frac{k}{x}\right] \leq \sum_{k=1}^{15} k }

Putting the limit \mathrm{ x \rightarrow 0^{+}, \: we \: have\: 120<L \leq 120 }

As the limit from both sides approaches to 120 , we have by sandwich theorem, the required limit \mathrm{ =120. }
 

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