For each x ε R, let [x] be the greatest integer less

Question:

For each $\mathrm{x} \varepsilon \mathrm{R}$, let $[\mathrm{x}]$ be the greatest integer less than or equal to $x$. Then

$\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to

  1. $-\sin 1$

  2. 0

  3. 1

  4. $\sin 1$


Correct Option: 1

Solution:

$\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$

$\mathrm{x} \rightarrow 0^{-}$

$[x]=-1 \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{x(-x-1) \sin (-1)}{-x}=-\sin 1$

$|x|=-x$

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