If [x] denotes the greatest integer less than | vedclass

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(C) Let $f(x) = 1 - x + [x - 1] + [1 - x]$.We evaluate the left-hand limit as $x 	o 1^-$:$\mathop {\lim }\limits_{x 	o 1^-} f(x) = \mathop {\lim }\l

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$-1$

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(C) Let $f(x) = 1 - x + [x - 1] + [1 - x]$.We evaluate the left-hand limit as $x 	o 1^-$:$\mathop {\lim }\limits_{x 	o 1^-} f(x) = \mathop {\lim }\limits_{h 	o 0^+} (1 - (1 - h) + [1 - h - 1] + [1 - (1 - h)])$$= \mathop {\lim }\limits_{h 	o 0^+} (h + [-h] + [h]) = \mathop {\lim }\limits_{h 	o 0^+} (h - 1 + 0) = -1$.We evaluate the right-hand limit as $x 	o 1^+$:$\mathop {\lim }\limits_{x 	o 1^+} f(x) = \mathop {\lim }\limits_{h 	o 0^+} (1 - (1 + h) + [1 + h - 1] + [1 - (1 + h)])$$= \mathop {\lim }\limits_{h 	o 0^+} (-h + [h] + [-h]) = \mathop {\lim }\limits_{h 	o 0^+} (-h + 0 - 1) = -1$.Since the left-hand limit equals the right-hand limit,the limit exists and is equal to $-1$.

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\mathop {\lim }\limits_{x \to 1} (1 - x + [x - 1] + [1 - x])$ is

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