Answer
$\lim\limits_{h\to 0}\frac{(-5+h)^2-25}{h}=-10$
Work Step by Step
$\lim\limits_{h\to 0}\frac{(-5+h)^2-25}{h}$
$=\lim\limits_{h\to 0}\frac{(25+h^2-10h)-25}{h}$ (we use $(a-b)^2=a^2-2ab+b^2$)
$=\lim\limits_{h\to 0}\frac{h^2-10h}{h}$
$=\lim\limits_{h\to 0}\frac{h(h-10)}{h}$ (factorize the numerator)
$=\lim\limits_{h\to 0}(h-10)$ (divide both the numerator and denominator by $h$)
$=0-10$
$=-10$
