$\lim _{x \rightarrow 0} \frac{(x+8)^{\frac{1}{3}}-2}{x}$

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 ：\(\boxed{\frac{1}{12}}\)