For each x in mathbb{R},let [x] represent the | vedclass

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(C) Given the limit: $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$.For $x \rightarrow 0^{-}$,we have $[x] = -1$ and $|x| = -x$.Substit

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$\sin 1$

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(C) Given the limit: $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$.For $x \rightarrow 0^{-}$,we have $[x] = -1$ and $|x| = -x$.Substituting these values into the expression:$\lim _{x \rightarrow 0^{-}} \frac{x(-1 + (-x)) \sin(-1)}{-x}$$= \lim _{x \rightarrow 0^{-}} \frac{x(-1 - x)(-\sin 1)}{-x}$$= \lim _{x \rightarrow 0^{-}} (1 + x) \sin 1$As $x \rightarrow 0$,this becomes $(1 + 0) \sin 1 = \sin 1$.

For each $x \in \mathbb{R}$,let $[x]$ represent the greatest integer function. Then $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to

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