1ppose you know that the derivatives of \( x^{\frac{1}{3}} \) is \( \frac{1}{3} x^{-\frac{2}{3}} \) for every \( x \neq 0 \). The \[ \lim _{x \rightarrow 8} \frac{\left(\frac{x}{8}\right)^{\frac{1}{3}}-1}{x-8}=\frac{1}{a} \text { where } a= \]

To solve this limit, we can apply the definition of the derivative. We want to find the limit as \( x \) approaches 8 of the expression: \[ \lim _{x \rightarrow 8} \frac{\left(\frac{x}{8}\right)^{\frac{1}{3}}-1}{x-8}. \] Setting \( u = \frac{x}{8} \), we have \( x = 8u \) and as \( x \to 8 \), \( u \to 1 \). The limit then transforms into: \[ \lim _{u \rightarrow 1} \frac{u^{\frac{1}{3}}-1}{8(u-1)}. \] The expression simplifies further, recognizing that the derivative of \( u^{\frac{1}{3}} \) at \( u = 1 \) is \( \frac{1}{3}(1)^{-\frac{2}{3}} = \frac{1}{3} \). Hence: \[ \lim _{u \rightarrow 1} \frac{u^{\frac{1}{3}}-1}{u-1} = \frac{1}{3}. \] Multiplying this by 8 gives: \[ \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{24}. \] Thus, equating \( \frac{1}{a} = \frac{1}{24} \) leads us to conclude that \( a = 24 \).