Answer
$$\lim _{h \rightarrow 0} \frac{(3+h)^3-27}{h}=27$$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{(3+h)^3-27}{h}$$
let $$ f(h) = \frac{(3+h)^3-27}{h}$$
Since, we have
$$ f(0)=\frac{27-27}{0}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{h \rightarrow 0} \frac{(3+h)^3-3}{h}\\
&= \lim _{h \rightarrow 0} \frac{27+9h^2+27h+h^3-27}{h}\\
&= \lim _{h \rightarrow 0} \frac{9h^2+27h+h^3}{h}\\
&= \lim _{h \rightarrow 0} \frac{h(9h+27+h^2)}{h}\\
&= \lim _{h \rightarrow 0}(9h+27+h^2)\\
&=0+27+0\\
&= 27
\end{aligned}
