Topic: Engineering Mathematics
The value of $\lim_{x\rightarrow8}\left(\frac{x^{1/3}-2}{x-8}\right)$ is
- A. $\frac{1}{12}$
- B. $\frac{1}{4}$
- C. $\frac{1}{16}$
- D. $\frac{1}{8}$
Answer: A. $\frac{1}{12}$
Explanation: This is a limit of the form $\frac{0}{0}$ as $x \to 8$.
Using L'Hôpital's rule, differentiate the numerator and the denominator:
$\lim_{x\to 8} \frac{\frac{d}{dx}(x^{1/3}-2)}{\frac{d}{dx}(x-8)} = \lim_{x\to 8} \frac{\frac{1}{3}x^{-2/3}}{1} = \frac{1}{3}(8)^{-2/3} = \frac{1}{3}(2^3)^{-2/3} = \frac{1}{3}(2^{-2}) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.