## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{h\to 0}\frac{(2+h)^3-8}{h}=12$
$\lim\limits_{h\to 0}\frac{(2+h)^3-8}{h}$ $=\lim\limits_{h\to 0}\frac{(8+h^3+12h+6h^2)-8}{h}$ [use $(a+b)^3=a^3+3a^2b+3ab^2+b^3$] $=\lim\limits_{h\to 0}\frac{h^3+6h^2+12h}{h}$ $=\lim\limits_{h\to 0}\frac{h(h^2+6h+12)}{h}$ (factorize the numerator) $=\lim\limits_{h\to 0}(h^2+6h+12)$ (divide both the numerator and denominator by $h$) $=(0^2+6\times0+12)$ $=12$