Question:
For each \(t\in R\), let \([t]\) be the greatest integer less than or equal to \(t\). Then \(\lim_{x\to0^+}x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\cdots+\left[\frac{15}{x}\right]\right)\)
For each \(t\in R\), let \([t]\) be the greatest integer less than or equal to \(t\). Then \(\lim_{x\to0^+}x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\cdots+\left[\frac{15}{x}\right]\right)\)
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Use \(t-1\lt[t]\le t\) for each term and squeeze.
Updated On: Nov 11, 2026
- does not exist in R
- is equal to 0
- is equal to 15
- is equal to 120
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The Correct Option is D
Solution and Explanation
For every real \(t\), \(t-1\lt[t]\le t\). So \(\frac{k}{x}-1\lt\left[\frac{k}{x}\right]\le\frac{k}{x}\) for \(k=1,\dots,15\).
Adding: \(\frac{1+2+\cdots+15}{x}-15\lt S\le\frac{1+2+\cdots+15}{x}\), i.e. \(\frac{120}{x}-15\lt S\le\frac{120}{x}\).
Multiply by \(x\gt 0\): \(120-15x\lt xS\le120\).
As \(x\to0^+\), the left side tends to 120, so by the squeeze theorem the limit is 120.
Answer: (4) is equal to 120
Adding: \(\frac{1+2+\cdots+15}{x}-15\lt S\le\frac{1+2+\cdots+15}{x}\), i.e. \(\frac{120}{x}-15\lt S\le\frac{120}{x}\).
Multiply by \(x\gt 0\): \(120-15x\lt xS\le120\).
As \(x\to0^+\), the left side tends to 120, so by the squeeze theorem the limit is 120.
Answer: (4) is equal to 120
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