## Thomas' Calculus 13th Edition

$2 \lt x \lt 8$
We notice that $g(x)=\dfrac{3}{x-2}=\dfrac{3}{3-[-(x-5)]}$ This implies that $(x) =\Sigma_{n=0}^{\infty} (\dfrac{-1}{3})^n (x-5)^n$ Remember that the series will converge when $|g(x)|=|\dfrac{3}{3-[-(x-5)]}| \lt 1$ This implies that $|\dfrac{x-5}{3 }| \lt 1$ or, $|x+5| \lt 3$ or, $-3 \lt x-5 \lt 5$ Thus, $2 \lt x \lt 8$