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JEE Main 20199 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 9 January · Evening Shift · Mathematics · Limits

For each x ∈ R , let [x] be the greatest integer less than or equal...

For each x∈Rx \in R, let [x]\left[x\right] be the greatest integer less than or equal to xx. Then lim⁡x→0−x([x]+∣x∣)sin⁡[x]∣x∣\lim_{x \to {0}^{-}}\dfrac{x\left(\left[x\right] + \left|x\right|\right)\sin \left[x\right]}{\left|x\right|} is equal to

Options

  1. A11
  2. B00
  3. C−sin⁡1- \sin 1
  4. Dsin⁡1\sin 1

Correct answer

C. −sin⁡1- \sin 1

Step-by-step solution

=lim⁡x→0−x([x]+∣x∣)sin⁡[x]∣x∣= \lim_{x \to {0}^{-}} \dfrac{x\left(\left[x\right] + \left|x\right|\right)\sin \left[x\right]}{\left|x\right|}

=lim⁡x→0−x(−1−x)sin⁡(−1)−x= \lim_{x \to {0}^{-}}\dfrac{x\left(- 1 - x\right)\sin \left(- 1\right)}{- x} \left\{\left[x\right] = - 1, \forall x \in [ - 1, 0 )\right\}<br/><br/><br /><br />= \lim_{x \to {0}^{-}} - \left(1 + x\right)\sin 1{sin⁡(−θ)=−sinθ}\left\{\sin \left(- \theta\right) = - sin\theta\right\}

=−sin⁡1= - \sin 1.
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