Answer
$4$
Work Step by Step
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$
