Answer
(a) (i) $\lim\limits_{x \to 2^+}g(x) = 5$
(ii) $\lim\limits_{x \to 2^-}g(x) = -5$
(b) $\lim\limits_{x \to 2}g(x)$ does not exist.
(c) We can see a sketch of the graph of $g(x)$ below.
Work Step by Step
(a) $g(x) = \frac{x^2+x-6}{\vert x-2\vert}$
(i) Note that $x \gt 2$
We can find $\lim\limits_{x \to 2^+}g(x)$:
$\lim\limits_{x \to 2^+}g(x)$
$=\lim\limits_{x \to 2^+}\frac{x^2+x-6}{\vert x-2\vert}$
$=\lim\limits_{x \to 2^+}\frac{(x+3)(x-2)}{x-2}$
$=\lim\limits_{x \to 2^+}(x+3)$
$= 5$
(ii) Note that $x \lt 2$
We can find $\lim\limits_{x \to 2^-}g(x)$:
$\lim\limits_{x \to 2^-}g(x)$
$=\lim\limits_{x \to 2^-}\frac{x^2+x-6}{\vert x-2\vert}$
$=\lim\limits_{x \to 2^-}\frac{(x+3)(x-2)}{-(x-2)}$
$=\lim\limits_{x \to 2^-}-(x+3)$
$= -5$
(b) Since $\lim\limits_{x \to 2^+}g(x)\neq \lim\limits_{x \to 2^-}g(x)$, then $\lim\limits_{x \to 2}g(x)$ does not exist.
(c) We can see a sketch of the graph of $g(x)$ below.
