Answer
$$2$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - 4{x^2} + 6x}}{{3x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}0{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 0} \left( {{x^3} - 4{x^2} + 6x} \right) = {\left( 0 \right)^3} - 4{\left( 0 \right)^2} + 6\left( 0 \right) = 0 \cr
& \mathop {\lim }\limits_{x \to 0} 3x = 3\left( 0 \right) = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{.Differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{D_x}\left( {{x^3} - 4{x^2} + 6x} \right)}}{{{D_x}\left( {3x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{3{x^2} - 8x + 6}}{3} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{{3{{\left( 0 \right)}^2} - 8\left( 0 \right) + 6}}{3} \cr
& = \frac{6}{3} \cr
& = 2 \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - 4{x^2} + 6x}}{{3x}} = 2 \cr} $$