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.

According to Clairaut's Theorem (also known as Schwarz's Theorem on the equality of mixed partials), if the mixed second-order partial derivatives $f_{xy}$ and $f_{yx}$ exist and are continuous in a neighbourhood of a point $(x_0, y_0)$, then they are equal:
$f_{yx}(x_0, y_0) = f_{xy}(x_0, y_0)$.

Detailed analysis of options:

• $f_{yx} \ne f_{xy}$ contradicts Clairaut's theorem.
• $f_{yx}(x_0, y_0) = f_{xy}(x_0, y_0)$ is the precise statement of Clairaut's theorem.
• Strict inequality is incorrect.
• Strict inequality is incorrect.

The Taylor series expansion of a smooth function $f(x)$ centered at $x = c$ is defined by the power series:
$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x - c)^n = f(c) + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \dots$.
When $c = 0$, it reduces to the Maclaurin series.

Detailed analysis of options:

• Uses powers of $x$ rather than $(x-c)$ with center $c$.
• Incorrect negative sign on the first derivative term.
• $f(c) + \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots$ is the standard definition of Taylor's series centered at $c$.
• Incorrect alternating signs and powers of $x$.

Evaluation of the left-hand limit:
As $x \to 1^-$, $x \lt 1$. For values of $x$ close to 1 with $0 \lt x \lt 1$, we have $x^2 \lt 1$, which implies $x^2 - 1 \lt 0$.
By definition of absolute value, when $u \lt 0$, $|u| = -u$.