Question

Solve :
$\underset{h\to 0}{lim}\frac{{\left(6+h\right)}^2-36}{h}$

Answer 1

$\begin{array}{c}{lim}_{h\to 0}\frac{{\left(6+h\right)}^2-36}{h}\\ ={lim}_{h\to 0}\frac{36+12h+h^2-36}{h}\\ ={lim}_{h\to 0}\frac{h\left(12+h\right)}{h}\\ ={lim}_{h\to 0}\left(12+h\right)\\ =12\end{array}$