Question:
For each \(x\in\mathbf{R}\), let \([x]\) be the greatest integer less than or equal to \(x\). Then \(\lim_{x\to0^-}\frac{x\left([x]+|x|\right)\sin[x]}{|x|}\) is equal to :
For each \(x\in\mathbf{R}\), let \([x]\) be the greatest integer less than or equal to \(x\). Then \(\lim_{x\to0^-}\frac{x\left([x]+|x|\right)\sin[x]}{|x|}\) is equal to :
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For \(x\) slightly negative: \([x]=-1\) and \(|x|=-x\).
Updated On: Nov 5, 2026
- \(-\sin 1\)
- \(1\)
- \(\sin 1\)
- \(0\)
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The Correct Option is A
Solution and Explanation
As \(x\to0^-\): \([x]=-1\), \(|x|=-x\), \(\frac{x}{|x|}=-1\).
Expression \(=\frac{x}{|x|}\,([x]+|x|)\sin[x]=(-1)(-1-x)\sin(-1)=(1+x)(-\sin1)\).
Limit \(=-\sin 1\).
Answer: (1) \(-\sin 1\)
Expression \(=\frac{x}{|x|}\,([x]+|x|)\sin[x]=(-1)(-1-x)\sin(-1)=(1+x)(-\sin1)\).
Limit \(=-\sin 1\).
Answer: (1) \(-\sin 1\)
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