Answer
$$\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}=1$$
Work Step by Step
Given $$\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}$$
let $$ f(x) = \frac{3 x^{2}-4 x-4}{2 x^{2}-8}$$
Since, we have
$$ f(2)= \frac{12-8-4}{8-8}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{x \rightarrow 2} \frac{3 x^{2}-4 x-4}{2 x^{2}-8}\\
&=\lim _{x \rightarrow 2} \frac{(3x+2)(x-2)}{2( x^{2}-4)}\\
&=\lim _{x \rightarrow 2} \frac{(3x+2)(x-2)}{2( x+2)(x-2)}\\
&=\lim _{x \rightarrow 2} \frac{(3x+2) }{2 (x+2)}\\
&= \frac{ 6+2}{8}\\
&=1
\end{aligned}
