Answer
$2 \lt x \lt 8$
Work Step by Step
We have
$g(x)=\dfrac{3}{x-2}=\dfrac{3}{3-[-(x-5)]}$
or, $=\Sigma_{n=0}^{\infty} (\dfrac{-1}{3})^n (x-5)^n$
The series will converge when $|g(x)|=|\dfrac{3}{3-[-(x-5)]}| \lt 1$
So, $|\dfrac{x-5}{3 }| \lt 1$
$ \implies |x+5| \lt 3$
$ \implies -3 \lt x-5 \lt 5$
$\implies 2 \lt x \lt 8$
