b) \( \lim _{x \rightarrow 4} \frac{3 x^{2}-8 x-16}{2 x^{2}-9 x+4} \)

To evaluate the limit \( \lim _{x \rightarrow 4} \frac{3 x^{2}-8 x-16}{2 x^{2}-9 x+4} \), we first substitute \( x = 4 \): For the numerator: \[ 3(4)^2 - 8(4) - 16 = 3(16) - 32 - 16 = 48 - 32 - 16 = 0 \] For the denominator: \[ 2(4)^2 - 9(4) + 4 = 2(16) - 36 + 4 = 32 - 36 + 4 = 0 \] Since both the numerator and denominator evaluate to zero, we can apply L'Hôpital's Rule, which states that if you have the indeterminate form \( \frac{0}{0} \), you can take the derivative of the numerator and denominator: 1. The derivative of the numerator \( 3x^2 - 8x - 16 \) is: \[ 6x - 8 \] 2. The derivative of the denominator \( 2x^2 - 9x + 4 \) is: \[ 4x - 9 \] We now find the limit again using these derivatives: \[ \lim _{x \rightarrow 4} \frac{6x - 8}{4x - 9} \] Substituting \( x = 4 \): \[ \frac{6(4) - 8}{4(4) - 9} = \frac{24 - 8}{16 - 9} = \frac{16}{7} \] Thus, the limit is: \[ \boxed{\frac{16}{7}} \]