## Calculus (3rd Edition)

$$\lim _{h \rightarrow 0} \frac{(3+h)^3-27}{h}=27$$
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}