Question

The value of $\underset{x\to 1}{lim}x\text{sgn}(x-1)$ is
(where 'sgn' represents signum function)

• A. $-1$
• B. $1$
• C. $0$
• D. limit does not exist

Answer 1

The correct option is D limit does not exist

Given : $\underset{x\to 1}{lim}x\text{sgn}(x-1)$
we know that
$\text{sgn}\left(x\right)=\{\begin{array}{cc}-1& x<0\\ 0& x=0\\ 1& x>0\end{array}$

$L.H.L.=\underset{x\to 1^{-}}{lim}x\text{sgn}(x-1)=\underset{h\to 0}{lim}(1-h)\text{sgn}(1-h-1)$
$=\underset{h\to 0}{lim}(1-h)\text{sgn}(-h)=1\times (-1)$
$=-1$
$R.H.L.=\underset{x\to 1^{+}}{lim}x\text{sgn}(x-1)=\underset{h\to 0}{lim}(1+h)\text{sgn}(1+h-1)$
$=\underset{h\to 0}{lim}(1+h)\text{sgn}\left(h\right)=1\times 1$
$=1$
$\Rightarrow L.H.L.\ne R.H.L.$
$\therefore$ Limit does not exist.