Answer
$$\lim _{h \rightarrow 0} \frac{(1+h)^3-1}{h}=3$$
Work Step by Step
Given $$\lim _{h \rightarrow 0} \frac{(1+h)^3-1}{h}$$
let $$ f(h) = \frac{(1+h)^3-1}{h}$$
Since, we have
$$ f(0)=\frac{1-1}{0}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{h \rightarrow 0} \frac{(1+h)^3-1}{h}\\
&= \lim _{h \rightarrow 0} \frac{1+3h^2+3h+h^3-1}{h}\\
&= \lim _{h \rightarrow 0} \frac{3h^2+3h+h^3}{h}\\
&= \lim _{h \rightarrow 0} \frac{h(3h+3+h^2)}{h}\\
&= \lim _{h \rightarrow 0}(3h+3+h^2)\\
&=0+3+0\\
&= 3
\end{aligned}
