Find the following limits. \( \lim _{x \rightarrow 2} \frac{8-x^{3}}{x^{3}-4 x}= \) a) \( =12] \)

To solve the limit \( \lim _{x \rightarrow 2} \frac{8-x^{3}}{x^{3}-4 x} \), start by substituting \( x = 2 \): \[ 8 - (2)^3 = 8 - 8 = 0 \] \[ (2)^3 - 4(2) = 8 - 8 = 0 \] Since substituting \( x = 2 \) gives the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can take the derivatives of the numerator and denominator. The derivative of the numerator \( 8 - x^3 \) is \( -3x^2 \). The derivative of the denominator \( x^3 - 4x \) is \( 3x^2 - 4 \). Now we can rewrite the limit: \[ \lim _{x \rightarrow 2} \frac{-3x^2}{3x^2 - 4} \] Substituting \( x = 2 \): \[ = \frac{-3(2)^2}{3(2)^2 - 4} = \frac{-3 \cdot 4}{3 \cdot 4 - 4} = \frac{-12}{12 - 4} = \frac{-12}{8} = -\frac{3}{2} \] Thus, the limit is \[ \lim _{x \rightarrow 2} \frac{8-x^{3}}{x^{3}-4 x} = -\frac{3}{2}. \] Therefore, the correct answer to the limit is \(-\frac{3}{2}\), not \(12\).