Question

Discuss the continuity of the $f\left(x\right)$ at the indicated points:
$f\left(x\right)=|x-1|+|x+1|$ at $x=-1,1$.

Answer 1

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

LHL at $x=-1$,

$\underset{x\to -1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(-1-h)=\underset{h\to 0}{lim}[|(-1-h)-1|+|(-1-h)+1|]=2+0=2$

RHL at $x=-1$,

$\underset{x\to -1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(-1+h)=\underset{h\to 0}{lim}[|(-1+h)-1|+|(-1+h)+1|]=2+0$

Also,

$f(-1)=|-1-1|+|-1+1|=2+0=2$

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

Now,

LHL at $x=1$,

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1-h)=\underset{h\to 0}{lim}[|(1-h)-1|+|(1-h)+1|]=0+2=2$

RHL at $x=1$,

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(1+h)=\underset{h\to 0}{lim}[|(1+h)-1|+|(1+h)+1|]=0+2=2$

Also,

$f\left(1\right)=|1-1|+|1+1|=0+2=2$

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

Hence $f\left(x\right)$ is continuous at $x=-1$ and $x=1$.