Let $f(x, y) = \ln u = \frac{x^3 + y^3}{3x + 4y}$.

• Testing homogeneity of $f(x, y)$:
  $$f(tx, ty) = \frac{(tx)^3 + (ty)^3}{3(tx) + 4(ty)} = \frac{t^3 (x^3 + y^3)}{t(3x + 4y)} = t^2 f(x, y)$$

Basic calculus limit property:
Since $r$ is a positive rational number ($r \gt 0$), as $x \to +\infty$, $x^r \to +\infty$.
For any finite constant $A \in \mathbb{R}$:

$$\lim_{x \to +\infty} \frac{A}{x^r} = \frac{A}{\infty} = 0$$

Expand $\sin 2x$ using Maclaurin series:

$$\sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \dots = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4}{3}x^3 + O(x^5)$$

Substitute into the given expression:

$$\frac{\sin 2x + ax + bx^3}{x^3} = \frac{(2x - \frac{4}{3}x^3 + O(x^5)) + ax + bx^3}{x^3}$$ $$= \frac{(2 + a)x + (b - \frac{4}{3})x^3 + O(x^5)}{x^3} = \frac{2 + a}{x^2} + \left(b - \frac{4}{3}\right) + O(x^2)$$

For the limit as $x \to 0$ to exist and be finite:

• The coefficient of $\frac{1}{x^2}$ must be zero:
  $$2 + a = 0 \implies a = -2$$
• The remaining constant term must equal the limit value 1:
  $$b - \frac{4}{3} = 1 \implies b = 1 + \frac{4}{3} = \frac{7}{3}$$

Standard calculus definition of a horizontal asymptote:
A horizontal line $y = L$ is defined as a horizontal asymptote of the graph of $y = f(x)$ if at least one of the following end-behavior limits holds:

$$\lim_{x \to +\infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L$$

As $x$ grows arbitrarily large in either positive or negative direction, the vertical distance between the curve $y = f(x)$ and the line $y = L$ approaches zero.