Answer
$$ - 4$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^3} - 7{x^2} + 12x}}{{4 - x}} \cr
& {\text{Try to evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^3} - 7{x^2} + 12x}}{{4 - x}} = \frac{{{{\left( 4 \right)}^3} - 7{{\left( 4 \right)}^2} + 12\left( 4 \right)}}{{4 - 4}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{64 - 112 + 48}}{{4 - 4}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{0}{0} \cr
& {\text{Factor the numerator}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^3} - 7{x^2} + 12x}}{{4 - x}} = \mathop {\lim }\limits_{x \to 4} \frac{{x\left( {{x^2} - 7x + 12} \right)}}{{4 - x}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 4} \frac{{x\left( {x - 4} \right)\left( {x - 3} \right)}}{{4 - x}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 4} \frac{{x\left( {4 - x} \right)\left( {3 - x} \right)}}{{4 - x}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 4} x\left( {3 - x} \right) \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{x \to 4} x\left( {3 - x} \right) = 4\left( {3 - 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 4 \cr} $$
