Answer
$$\lim _{x \rightarrow 3} \frac{x^{3}-27}{x^{2}-9}= \frac{9 }{2} $$
Work Step by Step
Given $$\lim _{x \rightarrow 3} \frac{x^{3}-27}{x^{2}-9}$$
let $$ f(x) = \lim _{x \rightarrow 3} \frac{x^{3}-27}{x^{2}-9}$$
Since, we have
$$ f( 3)= \lim _{x \rightarrow 3} \frac{27-27}{9-9}=\frac{0}{0}$$
So, transform algebraically and cancel, we get
\begin{aligned}
L&=\lim _{x \rightarrow 3} \frac{x^{3}-27}{x^{2}-9}\\
&= \lim _{x \rightarrow 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}\\
&= \lim _{x \rightarrow 3} \frac{ (x^2+3x+9)}{ (x+3)}\\
&= \frac{9+9+9 }{3+3}\\
&= \frac{27 }{6} \\
&=\frac{9 }{2}
\end{aligned}
