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JEE MATHEMATICS MCQ: LIMITS

1 Easy
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Analyzing +4 | -1
For each t∈Rt \in R, let [t] denote the greatest integer less than or equal to t. Then,
lim⁡x→1+(1−∣x∣+sin⁡∣1−x∣)sin⁡(π2[1−x])∣1−x∣[1−x]\lim \limits_ {x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}
View answer and explanation
Correct option: D
Equals 0
Option D (Equals 0) is correct because after simplifying the expression and evaluating the limit as x approaches 1 from the right, the result is 0.

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