Since the cube root function $g(u) = \sqrt[3]{u}$ is continuous on $\mathbb{R}$, we can pull the limit inside the root:
$\lim_{x \to +\infty} \sqrt[3]{\frac{3x+5}{6x-8}} = \sqrt[3]{\lim_{x \to +\infty} \frac{3x+5}{6x-8}}$.
Dividing numerator and denominator by $x$:
$\lim_{x \to +\infty} \frac{3 + 5/x}{6 - 8/x} = \frac{3 + 0}{6 - 0} = \frac{3}{6} = \frac{1}{2}$.
Taking the cube root:
$\sqrt[3]{\frac{1}{2}}$.

Detailed analysis of options:

• $\sqrt[3]{1/3}$ is incorrect.
• $\sqrt[3]{1/5}$ is incorrect.
• $\sqrt[3]{1/7}$ is incorrect.
• $\sqrt[3]{\frac{1}{2}}$ is the correct answer.