JEE PYQ: Limits Question 29
Question 29 - 2019 (09 Jan Shift 2)
For each $x \in \mathbb{R}$, let $[x]$ be greatest integer less than or equal to $x$. Then $\lim_{x \to 0^-} \frac{x([x] + |x|)\sin[x]}{|x|}$ is equal to:
(1) $-\sin 1$
(2) $1$
(3) $\sin 1$
(4) $0$
Show Answer
Answer: (1)
Solution
Let $x = 0 - h$, $h > 0$. $= \lim_{h\to 0}\frac{(-h)([-h]+|-h|)\sin[-h]}{|-h|} = \lim_{h\to 0}\frac{(-h)(-1+h)\sin(-1)}{h} = (0-1)(-1)\sin(-1) \cdot (-1) = -\sin 1$.
BETA
