Question

If $f\left(x\right)=\{\begin{array}{cc}|x|+1,& x<0\\ 0,& x=0\\ |x|-1,& x>0\end{array}$ For what value(s) of $a$ does $\underset{x\to a}{lim}f\left(x\right)$ exists ?

Answer 1

Step $1:$ Simplification given data
Given $f\left(x\right)=\{\begin{array}{cc}|x|+1,& x<0\\ 0,& x=0\\ |x|-1,& x>0\end{array}$

We check limit different values of $a$
When $a=0$
When $a<0$
When $a>0$

Step $2:$ Verification of different cases

Case $1:$ When $a=0$
Limit exists at $a=0$ if $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$f\left(x\right)=\{\begin{array}{cc}|x|+1,& x<0\\ 0,& x=0\\ |x|-1,& x>0\end{array}$
$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}f(-h)$
$=\underset{h\to 0}{lim}(|-h|+1)$
$=\underset{h\to 0}{lim}(h+1)$
$=0+1$
$=1$

$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}f\left(h\right)$
$=\underset{h\to 0}{lim}(|h|-1)$
$=\underset{h\to 0}{lim}(h-1)$
$=0-1$
$=-1$
Since $L.H.L.\ne R.H.L.$
$\therefore$ At $x=0,$ Limit does not exist.

Case $2:$ When $a<0$
For $a<0$
$f\left(x\right)=|x|+1$
$f\left(x\right)=-x+1$ ( As $x$ is negative )
Since this a polynomial
Limit exists at every point less than $0$.
$\therefore$ Limit exists for $a<0$.

Case $3:$ When $a>0$
For $a>0$
$f\left(x\right)=|x|-1$
$f\left(x\right)=x+1$ ( As $x$ is positive )
Since this a polynomial
Limit exists at every point greater than $0$
$\therefore$ Limit exists for $a>0$

$\therefore \underset{x\to a}{lim}f\left(x\right)$ exists for all values of $a$, where $a\ne 0$