JEE PYQ: Limits Question 30
Question 30 - 2019 (10 Jan Shift 1)
For each $t \in \mathbb{R}$, let $[t]$ be the greatest integer less than or equal to $t$. Then, $\lim_{x \to 1^+} \frac{(1 - |x| + \sin|1-x|)\sin\left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}$
(1) equals $1$
(2) equals $0$
(3) equals $-1$
(4) does not exist
Show Answer
Answer: (2)
Solution
Let $x = 1 + h$, $h \to 0^+$. Numerator: $(1 - 1 - h + \sin h)\sin\left(\frac{\pi}{2}(-1)\right) = (-h + \sin h)\sin\left(-\frac{\pi}{2}\right)$. Denominator: $h \cdot (-1)$. $= \lim_{h\to 0}\frac{(-h + \sin h)(-1)}{-h} = \lim_{h\to 0}\frac{h - \sin h}{h} = 0$.
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