Answer
$$\frac{1}{{12}}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}8{\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to 8} \left( {\root 3 \of x - 2} \right) = \root 3 \of 8 - 2 = 2 - 2 = 0 \cr
& \mathop {\lim }\limits_{x \to 8} \left( {x - 8} \right) = 8 - 8 = 0 \cr
& \cr
& {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. Differentiating the numerator and}} \cr
& {\text{denominator}}{\text{}} \cr
& {\text{for }}\root 3 \of x - 2 \to {D_x}\left( {{x^{1/3}} - 2} \right) = \frac{1}{3}{x^{ - 2/3}} - 0 = \frac{1}{{3\root 3 \of {{x^2}} }} \cr
& {\text{for }}\left( {x - 8} \right) \to {D_x}\left( {x - 8} \right) = 1 \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} = \mathop {\lim }\limits_{x \to 8} \frac{1}{{3\root 3 \of {{x^2}} }} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \frac{1}{{3\root 3 \of {{{\left( 8 \right)}^2}} }} \cr
& = \frac{1}{{3\left( 4 \right)}} \cr
& = \frac{1}{{12}} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to 8} \frac{{\root 3 \of x - 2}}{{x - 8}} = \frac{1}{{12}} \cr} $$
