limx→1 ((1−x)+[x−1]+|1−x|) where ∣[x] denotesthe greatest integer less than or equal to x

# $\underset{x\to 1}{lim} \left(\left(1-x\right)+\left[x-1\right]+|1-x|\right)$ where $\mid \left[x\right]$ denotesthe greatest integer less than or equal to x

1. A

is equal to 0

2. B

is equal to 1

3. C

does not exist

4. D

is equal to –1

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### Solution:

For $0, so

$F\left(x\right)=\left(1-x\right)+\left(-1\right)+\left(1-x\right)=1-2x$

$⇒\underset{x\to 1-}{lim} f\left(x\right)=1-2.\left(1\right)=-1$.

For $1, so

$f\left(x\right)=\left(1-x\right)+0+\left(x-1\right)=0$

$⇒\underset{x\to 1+}{lim} f\left(x\right)=0$ Thus $\underset{x\to ||}{lim} f\left(x\right)$ does not exist.

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