\( 5 . \lim _ { x \rightarrow 2 } \frac { x ^ { 4 } + x ^ { 3 } - 24 } { x ^ { 2 } - 6 } = \frac { 12 } { } \)

To solve the limit, let's first calculate the expression inside the limit as \( x \) approaches 2. We start with substituting \( x = 2 \): 1. The numerator: \[ (2^4) + (2^3) - 24 = 16 + 8 - 24 = 0 \] 2. The denominator: \[ (2^2) - 6 = 4 - 6 = -2 \] Since both the numerator and denominator tend towards 0, we can apply L'Hôpital's Rule. We'll differentiate the numerator and the denominator separately: - Derivative of the numerator: \[ \frac{d}{dx}(x^4 + x^3 - 24) = 4x^3 + 3x^2 \] - Derivative of the denominator: \[ \frac{d}{dx}(x^2 - 6) = 2x \] Now we evaluate the limit again: \[ \lim_{x \rightarrow 2} \frac{4x^3 + 3x^2}{2x} \] Substituting \( x = 2 \): - Numerator: \[ 4(2^3) + 3(2^2) = 4(8) + 3(4) = 32 + 12 = 44 \] - Denominator: \[ 2(2) = 4 \] So we have: \[ \lim_{x \rightarrow 2} \frac{4x^3 + 3x^2}{2x} = \frac{44}{4} = 11 \] Now, multiplying by 5: \[ 5 \cdot 11 = 55 \] Thus, the value of the expression is: \[ 5 \cdot \lim _ { x \rightarrow 2 } \frac { x ^ { 4 } + x ^ { 3 } - 24 } { x ^ { 2 } - 6 } = 55 \] This means that the statement \( 5 \cdot \lim _ { x \rightarrow 2 } \frac { x ^ { 4 } + x ^ { 3 } - 24 } { x ^ { 2 } - 6 } = \frac { 12 }{ } \) is incorrect. The correct value is 55.