Answer
$\lim\limits_{h \to 0}\frac{(h-5)^2-25}{h}=-10$
Work Step by Step
$\lim\limits_{h \to 0}\frac{(h-5)^2-25}{h}=\lim\limits_{h \to 0}\frac{h^2-10h+25-25}{h}=\lim\limits_{h \to 0}\frac{h^2-10h}{h}=\lim\limits_{h \to 0}\frac{h(h-10)}{h}=\lim\limits_{h \to 0}\frac{(h-10)}{1}=\lim\limits_{h \to 0}{(h-10)}=-10$
