We are tasked with finding the values of $a$ and $b$ such that:

$\lim_{x \to 0} \frac{x \left( 1 - a \cos x \right) + b \sin 2x}{x^3} = 2$

Step 1: Expand $\cos x$ and $\sin 2x$ using their Taylor series approximations

1. Cosine series: $\cos x = 1 - \frac{x^2}{2} + O(x^4)$

2. Sine series: $\sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5)$

Step 2: Substitute these expansions into the expression

Substitute the Taylor series of $\cos x$ and $\sin 2x$ into the original expression:

$x \left( 1 - a \cos x \right) + b \sin 2x = x \left( 1 - a \left( 1 - \frac{x^2}{2} + O(x^4) \right) \right) + b \left( 2x - \frac{4x^3}{3} + O(x^5) \right)$

Now simplify:

$= x \left( 1 - a + a \frac{x^2}{2} + O(x^4) \right) + b \left( 2x - \frac{4x^3}{3} + O(x^5) \right)$

$= x (1 - a) + \frac{a x^3}{2} + 2bx - \frac{4b x^3}{3} + O(x^5)$

Combine like terms:

$= (1 - a + 2b) x + \left( \frac{a}{2} - \frac{4b}{3} \right) x^3 + O(x^5)$

Step 3: Divide by $x^3$

Now divide the entire expression by $x^3$:

$\frac{(1 - a + 2b) x + \left( \frac{a}{2} - \frac{4b}{3} \right) x^3 + O(x^5)}{x^3}$

$= \frac{1 - a + 2b}{x^2} + \left( \frac{a}{2} - \frac{4b}{3} \right) + O(x^2)$

Step 4: Analyze the limit as $x \to 0$

For the limit to exist and equal 2, the coefficient of $\frac{1}{x^2}$ must be zero (otherwise, the limit would tend to infinity), and the constant term must be equal to 2.

From $\frac{1 - a + 2b}{x^2} = 0$, we get:

$1 - a + 2b = 0 \quad \text{(Equation 1)}$

From the constant term, we have:

$\frac{a}{2} - \frac{4b}{3} = 2 \quad \text{(Equation 2)}$

Step 5: Solve the system of equations

1. From Equation 1, we get: $a = 1 + 2b$

2. Substitute $a = 1 + 2b$ into Equation 2:

$\frac{1 + 2b}{2} - \frac{4b}{3} = 2$

Multiply through by 6 to eliminate fractions:

$3(1 + 2b) - 8b = 12$

Simplify:

$3 + 6b - 8b = 12$

$3 - 2b = 12$

$-2b = 9 \quad \Rightarrow \quad b = -\frac{9}{2}$

3. Substitute $b = -\frac{9}{2}$ into $a = 1 + 2b$:

$a = 1 + 2 \times \left(-\frac{9}{2}\right) = 1 - 9 = -8$

Final Answer:

The values of $a$ and $b$ are:

$a = -8, \quad b = -\frac{9}{2}$

Would you like further explanation or more examples? Here are some related questions:

1. What are the Taylor series expansions of higher-order terms for $\sin x$ and $\cos x$?
2. How can we apply the method of series expansions to other limits?
3. What happens if we change the limit to a different power of $x$?
4. How would the solution change if the denominator was $x^2$ instead of $x^3$?
5. Can you use the same method to find limits of trigonometric expressions involving more terms?

Tip: Always look for dominant terms in the numerator when solving limits involving series expansions.