Given that lim (x → 0) [(1 - a cos 3x) + 6 sin x] / [sin x + a x + b x^3] is finite, find the values of a and b.

Steps to Solve

1. Identify the Limit Expression
   We need to analyze the limit:
   $$ \lim_{x \to 0} \frac{(1 - a \cos 3x) + 6 \sin x}{\sin x + a x + b x^3} $$ for it to exist and be finite.

2. Evaluate the Numerator at x = 0
   Substituting $x = 0$ in the numerator:
   $$ 1 - a \cos(0) + 6 \sin(0) = 1 - a $$

3. Evaluate the Denominator at x = 0
   Substituting $x = 0$ in the denominator:
   $$ \sin(0) + a(0) + b(0)^3 = 0 $$
   Since the denominator approaches 0, we need the numerator to also approach 0 for the limit to be finite.

4. Set the Numerator to Zero
   For the limit to be finite, we must have:
   $$ 1 - a = 0 \implies a = 1 $$

5. Substitute a into the Limit Expression
   Now substitute $a = 1$ into the original limit:
   $$ \lim_{x \to 0} \frac{(1 - \cos 3x) + 6 \sin x}{\sin x + x + b x^3} $$

6. Simplify the New Limit Expression
   As $x \to 0$, we have:

• $1 - \cos(3x) \approx \frac{(3x)^2}{2} = \frac{9x^2}{2}$
• $6 \sin x \approx 6x$
  Thus, the numerator is approximately:
  $$ \frac{9x^2}{2} + 6x $$

7. Evaluate the New Denominator Close to x = 0
   Substituting $a = 1$ in the denominator gives:
   $$ \sin x + x + b x^3 \approx x + x + b x^3 = 2x + b x^3 $$

8. Formulate Limit Expression
   Now calculating the limit becomes:
   $$ \lim_{x \to 0} \frac{6x + \frac{9x^2}{2}}{2x + b x^3} $$
   This leads to:
   $$ \lim_{x \to 0} \frac{x(6 + \frac{9x}{2})}{x(2 + b x^2)} = \lim_{x \to 0} \frac{6 + \frac{9x}{2}}{2 + b x^2} $$

9. Evaluate the Limit as x Approaches 0
   Taking the limit as $x \to 0$:
   $$ \frac{6}{2} \implies \text{The limit exists and equals } 3 $$
   We want the limit to match the given finite condition, so we must have $b = 0$.