## Calculus 8th Edition

$-1.5$
From the given graph, we have $\lim\limits_{x \to 2} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to 2} \dfrac{1.5(x-2)}{(2-x)} = \dfrac{0}{0}$ This shows is an indeterminate form. Thus, applying L'Hospital, we get $= \lim\limits_{x \to 2} \dfrac{1.5(x-2)}{(-1)(x-2)}$ $=-1.5$