$\lim _{x \rightarrow 0} \frac{(x+8)^{\frac{1}{3}}-2}{x}$
\(\boxed{\frac{1}{12}}\)
Step 1 :Check if the given expression is in the indeterminate form (0/0 or ∞/∞) as x approaches 0:
Step 2 :\(\lim_{x \rightarrow 0} \frac{(x+8)^{\frac{1}{3}}-2}{x} = \frac{((0+8)^{\frac{1}{3}}-2)}{0} = \frac{2-2}{0} = \frac{0}{0}\)
Step 3 :Since the expression is in the indeterminate form, we can use L'Hopital's rule to find the limit. First, find the derivative of the numerator and denominator with respect to x:
Step 4 :\(\frac{d}{dx}((x+8)^{\frac{1}{3}}-2) = \frac{1}{3}(x+8)^{-\frac{2}{3}}\)
Step 5 :\(\frac{d}{dx}(x) = 1\)
Step 6 :Apply L'Hopital's rule and find the limit of the new expression:
Step 7 :\(\lim_{x \rightarrow 0} \frac{\frac{1}{3}(x+8)^{-\frac{2}{3}}}{1} = \frac{1}{3}(0+8)^{-\frac{2}{3}} = \frac{1}{3}(8^{-\frac{2}{3}}) = \frac{1}{3}(\frac{1}{4})\)
Step 8 :\(\boxed{\frac{1}{12}}\)