Let's carefully go through the solution using L'Hopital's Rule: \[\] The given problem is: \[ \lim _{x \rightarrow 1} \frac{4 x^3}{1} = \lim _{x \rightarrow k} \frac{3 x^2}{2x} \] We need to evaluate both limits. \[\] Left-hand side: \[\] We evaluate the limit as \(x \to 1\): \[ \lim _{x \rightarrow 1} \frac{4 x^3}{1} = 4 \times 1^3 = 4 \] Right-hand side: \[\] We now consider the second expression: \[ \lim _{x \rightarrow k} \frac{3 x^2}{2x} \] Simplify the expression: \[ \frac{3 x^2}{2x} = \frac{3x}{2} \] Thus, the limit becomes: \[ \lim _{x \rightarrow k} \frac{3 x}{2} = \frac{3k}{2} \] Equating both sides of the problem: \[ 4 = \frac{3k}{2} \] Now, solve for \( k \): \[ k = \frac{8}{3} \]