## Precalculus (6th Edition) Blitzer

$4$
Now, $m_{\tan}=\lim_\limits{ h\to 0} \dfrac{f(2+h)-f(2)}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{(2+h)^2-2^2}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{4+4h+h^2-4}{h}$ or, $=\lim_\limits{ h\to 0} (h+4)$ or, $=0+4$ or, $m_{tan}=4$