Answer
$4$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow 3} \frac{ 4x -12 }{x^2-5x+6}= \frac{0}{0}.
$$
Then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 3} \frac{ 4x -12 }{x^2-5x+6}=\lim _{x \rightarrow 3} \frac{ 4 }{2x-5}= { 4}{ 6-5}=4.
$$
