if $\underset{x\to 1}{lim} (2-x+a[x-1]+b[1+x\left]\right)$ exists, then a and b can take the value (where[.] denotes the greatest  integer function)

Since the greatest function is discontinuous (sensitive) at integral values of x, for a given limit to exist both left–and right–hand limits must be equal.

$\text{ L.H.L. }=\underset{x\to 1^{-}}{lim} (2-x+a[x-1]+b[1+x\left]\right)$

$=2-1+a(-1)+b\left(1\right)=1-a+b$

R.H.L.$=\underset{x\to 1^{+}}{lim} (2-x+a[x-1]+b[1+x\left]\right)$

$=2-1+a\left(0\right)+b\left(2\right)=1+2b$

On comparing, we have $–a = b$.