b) \( \lim _{x \rightarrow +\infty} \frac{(3 x+2)^{3}}{2 x(3 x+1)(4 x-1)}, \mathrm{R}: \frac{9}{8} \)

To evaluate the limit, we start by simplifying the expression inside the limit: \[ \lim _{x \rightarrow +\infty} \frac{(3 x+2)^{3}}{2 x(3 x+1)(4 x-1)} \] First, we expand the numerator: \[ (3x + 2)^3 = 27x^3 + 3 \cdot 3^2 \cdot 2x^2 + 3 \cdot 3 \cdot 2^2 x + 8 = 27x^3 + 54x^2 + 36x + 8 \] For large \(x\), the \(27x^3\) term dominates. Thus, we can approximate the numerator as: \[ (3x + 2)^3 \approx 27x^3 \] Now let's look at the denominator: \[ 2x(3x + 1)(4x - 1) = 2x(12x^2 - 3x + 4x - 1) = 2x(12x^2 + x - 1) \approx 24x^3 \] Now substituting our approximations back into the limit, we have: \[ \lim _{x \rightarrow +\infty} \frac{(3x + 2)^{3}}{2x(3x + 1)(4x - 1)} \approx \lim _{x \rightarrow +\infty} \frac{27x^3}{24x^3} \] This simplifies down to: \[ \frac{27}{24} = \frac{9}{8} \] Thus, the final result of the limit is: \[ \lim _{x \rightarrow +\infty} \frac{(3 x+2)^{3}}{2 x(3 x+1)(4 x-1)} = \frac{9}{8} \] This confirms the result you provided.