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For each $x \in R,$ let $[x]$ be the greatest integer less than or equal to $x .$ Then,
$lim_{x \to 1^{-}} \frac{x ([x] + | x |) \sin [x]}{| x |}$ is equal to

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$\sin 1$

$- \sin 1$

answer is A.

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Detailed Solution

$lim_{x \to 0^{-}} \frac{x ([x] + | x |) \sin [x]}{| x |}$

$= lim_{x \to 0^{-}} \frac{x (- 1 - x) \sin (- 1)}{- x} [\begin{array}{l} For - 1 < x < 0, \\ [x] = - 1, | x | = - x \end{array}]$

$= lim_{x \to 0^{-}} - (1 + x) \sin 1 = - \sin 1$

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