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Integration as Antiderivative

For each xε...

Question

For each $x ε R$ , let $[x]$ be the greatest integer less than or equal to $x$ .Then
$lim x \to 0^{-} \frac{x ([x] + | x |) sin [x]}{| x |}$ is equal to

- sin 1

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0

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1

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sin 1

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Solution

The correct option is B $- sin 1$
$lim x \to 0^{-} \frac{x ([x] + | x |) sin [x]}{| x |}$
When $x \to 0^{-}$
$[x] = - 1$
$| x | = - x$
$\Rightarrow lim x \to 0^{-} \frac{x (- x - 1) sin (- 1)}{- x} = - sin 1$

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