Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

For finding $\underset{x\to 1}{lim}(1-x+[x-1\left]\right[1-x\left]\right)$ we find $\underset{x\to 1^{+}}{lim}$ and ${lim}_{x\to 1^{-}}$:

Now, at $x\to 1^{+}$:

$[x-1]=0$ & $[1-x]=-1$ [Greatest integer function]

and at ${lim}_{x\to 1^{-}}$

$[x-1]=-1$ & $[1-x]=0$ [Greatest integer function]

So,

$\underset{x\to 1^{+}}{lim}(1-x+[x-1\left]\right[1-x\left]\right)$

$=\underset{x\to 1^{+}}{lim}(1-1^{+}+(0\left)\right(1\left)\right)$

$=0$

and

$\underset{x\to 1^{-}}{lim}(1-x+[x-1\left]\right[1-x\left]\right)$

$=\underset{x\to 1^{-}}{lim}(1-1^{-}+(1\left)\right(0\left)\right)$

$=0$

Hence as both RHL and LHL coincides, thus $\underset{x\to 1}{lim}(1-x+[x-1\left]\right[1-x\left]\right)=0$.

Hence, correct option is A.