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

The given function is
$f(x)=\left\{\begin{array}{1 1}|x|+1,&x<0\\0,&x=0 \\|x|-1,&x>0\end{array}\right.$
When $a = 0$
$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-}(|x|+1)$
$= \lim\limits_{x \to 0}(-x+1) \quad \quad [ If\: x < 0, \: |x|=-x]$
$= -0+1$
$= 1$
$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+}(|x|-1)$
$= \lim\limits_{x \to 0}(x-1) \quad \quad [ If\: x > 0, \: |x|=x]$
$= 0-1$
$= -1$
Here it is observed that $\lim\limits_{x \to 0^-} f(x) \neq \lim\limits_{x \to 0^+} f(x)$
$\therefore \lim\limits_{x \to 0} f(x)$ does not exist.
when $a < 0$
$\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^-}(|x|+1)$
$= \lim\limits_{x \to a}(-x+1) \quad \quad [ If\: x < a < 0 \Rightarrow |x|=-x]$
$= -a + 1$
$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+}(|x|+1)$
$= \lim\limits_{x \to a}(-x+1) \quad \quad [ If\: a < x < 0 \Rightarrow |x|=-x]$
$= -a + 1$
$\therefore \lim\limits_{x \to a^-}f(x) = \lim\limits_{x \to a^+}f(x) = -a+1$
Thus, the limit of $f(x)$ exists at $x = a$, where $a < 0$
When $a > 0$
$\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^-}(|x|-1)$
$= \lim\limits_{x \to a} (x-1) \qquad [0 < x < a \Rightarrow |x| = x]$
$= a-1$
$\lim\limits_{x \to a^+} f(x) = \lim\limits_{x \to a^+}(|x|-1)$
$= \lim\limits_{x \to a} (x-1) \qquad [0 < a < x \Rightarrow |x| = x]$
$= a-1$
$\therefore \lim\limits_{x \to a^-}f(x) = \lim\limits_{x \to a^+}f(x) = a-1$
Thus, the limit of $f(x)$ exists at $x = a$, where $a > 0$
Thus, $\lim\limits_{x \to a}f(x)$ exist for all $a \neq 0$