$\underset{x\to 1}{lim} \left(\right(1-x)+[x-1]+|1-x\left|\right)$ where $\mid \left[x\right]$ denotes

the greatest integer less than or equal to x

For $0<x<1,-1<x-1<0$, so

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

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

For $1<x<2,0<x-1<1$, so 

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

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