If $f(x)= \begin{cases}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{cases}$

For what value (s) of a does $\lim _{x \rightarrow a} f(x)$ exists?


The given function is

$f(x)= \begin{cases}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{cases}$

When $a=0$,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(|x|+1)$

$=\lim _{x \rightarrow 0}(-x+1) \quad[$ If $x<0,|x|=-x]$



$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(|x|-1)$

$=\lim _{x \rightarrow 0}(x-1) \quad[$ If $x>0,|x|=x]$



Here, it is observed that $\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$.

$\therefore \lim _{x \rightarrow 0} f(x)$ does not exist.

When $a<0$,

$\lim _{x \rightarrow d} f(x)=\lim _{x \rightarrow 0^{-}}(|x|+1)$

$=\lim _{x \rightarrow a}(-x+1) \quad[x


$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}}(|x|+1)$

$=\lim _{x \rightarrow a}(x-1) \quad[0


$\therefore \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=a-1$

Thus, $\lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$.

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