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$If f (x) = | x - 1 | - [x]$ (Where [X] is greatest integer less than or equal to x) then.

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$\lim_{x \to 1} f (x) exists$

$f (1^{+}) = 0 = f (1^{-})$

$f (1^{+}) = - 1; f (1^{-}) = 0$

answer is A.

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Detailed Solution

$\begin{matrix} f (1^{+}) = x - 1 - 1 \\ = x - 2 \\ = 1 - 2 \\ = - 1 \\ f (1^{-}) = - (x - 1) - 0 \\ = - x + 1 \\ = - 1 + 1 \\ = 0 \end{matrix}$

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