Question

If $f\left(x\right)=x\text{sgn}(x-1),$ then

• A. $f\left(x\right)$ is continuous at $x=1$
• B. $f\left(x\right)$ is discontinuous at $x=1$
• C. $\underset{x\to 1^{+}}{lim}x\text{sgn}(x-1)=-1$
• D. $\underset{x\to 1^{-}}{lim}x\text{sgn}(x-1)=-1$
  

Answer 1

Answer: (B), (D)

$\underset{x\to 1^{-}}{lim}x\text{sgn}(x-1)=-1$


Given : $f\left(x\right)=x\text{sgn}(x-1)$
we know,
$\text{sgn}\left(x\right)=\{\begin{array}{cc}-1& x<0\\ 0& x=0\\ 1& x>0\end{array}$
So, only critical point is $x=1$

$L.H.L.=\underset{x\to 1^{-}}{lim}x\text{sgn}(x-1)=\underset{h\to 0}{lim}(1-h)\text{sgn}(1-h-1)$
$=\underset{h\to 0}{lim}(1-h)\text{sgn}(-h)=1\times (-1)$
$=-1$
$R.H.L.=\underset{x\to 1^{+}}{lim}x\text{sgn}(x-1)=\underset{h\to 0}{lim}(1+h)\text{sgn}(1+h-1)$
$=\underset{h\to 0}{lim}(1+h)\text{sgn}\left(h\right)=1\times 1$
$=1$
$\Rightarrow L.H.L.\ne R.H.L.$
$\therefore$ $f\left(x\right)$ is discontinuous at $x=1.$