Answer
$$ \lim _{x \rightarrow 27} \frac{x-27}{x^{1 / 3}-3}=27$$
Work Step by Step
Given $$ \lim _{x \rightarrow 27} \frac{x-27}{x^{1 / 3}-3}$$
let $$ f(x) = \frac{x-27}{x^{1 / 3}-3}$$
Since, we have
$$ f( 27)= \frac{27-27}{3-3}=\frac{0}{0}$$
So, transform algebraically and cancel, we get
\begin{aligned}
L&= \lim _{x \rightarrow 27} \frac{x-27}{x^{1 / 3}-3}\\
&= \lim _{x \rightarrow 27} \frac{(x^\frac{1}{3})^3-3^3}{x^{1 / 3}-3}\\
&=\lim _{x \rightarrow 27} \frac{\left(x^{1 / 3}-3\right)\left(x^{2 / 3}+3 x^{1 / 3}+9\right)}{x^{1 / 3}-3}\\
&=\lim _{x \rightarrow 27}\left(x^{2 / 3}+3 x^{1 / 3}+9\right)\\
&=9+9+9\\
&=27
\end{aligned}
