If the function f(x, y) has mixed second order partial... | Agrigyan

When $x \to +\infty$ and $r \gt 0$, $x^r \to +\infty$.

By definition, the graph of a continuous function $f$ has a vertical tangent line at $x = c$ if the two-sided limit of the difference quotient (or derivative) is infinite with the same sign:
$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = +\infty$ (or $-\infty$).
This requires both one-sided limits to agree and tend to $+\infty$ (or both to $-\infty$).
When one side is $+\infty$ and the other is $-\infty$, the graph has a cusp (point of non-differentiability without a unique tangent line), not a vertical tangent line.

Detailed analysis of options:

• $\lim_{x \to c^-} f'(x) = +\infty$ and $\lim_{x \to c^+} f'(x) = +\infty$ is the definition of a vertical tangent.
• Opposite infinite limits define a cusp (e.g., $y = x^{2/3}$), not a tangent line.
• Opposite infinite limits define a cusp, not a tangent line.
• One finite and one infinite limit does not produce a vertical tangent.

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.