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:
  $$b - \frac{4}{3} = 1 \implies b = 1 + \frac{4}{3} = \frac{7}{3}$$