## Calculus with Applications (10th Edition)

$$\frac{1}{{27}}$$
\eqalign{ & \mathop {\lim }\limits_{x \to 27} \frac{{\root 3 \of x - 3}}{{x - 27}} \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting 2}}7{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 27} \left( {\root 3 \of x - 3} \right) = \root 3 \of {27} - 3 = 3 - 3 = 0 \cr & \mathop {\lim }\limits_{x \to 27} \left( {x - 27} \right) = 27 - 27 = 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 - 3 \to {D_x}\left( {\root 3 \of x - 3} \right) = \frac{1}{3}{x^{ - 2/3}} - 0 = \frac{1}{{3\root 3 \of {{x^2}} }} \cr & {\text{for }}\left( {x - 27} \right) \to {D_x}\left( {x - 27} \right) = 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 27} \frac{{\root 3 \of x - 3}}{{x - 27}} = \mathop {\lim }\limits_{x \to 27} \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( {27} \right)}^2}} }} \cr & = \frac{1}{{3\left( 9 \right)}} \cr & = \frac{1}{{27}} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 27} \frac{{\root 3 \of x - 3}}{{x - 27}} = \frac{1}{{27}} \cr}