Find the values of a and b so that \( \lim_{x \to 0} \frac{x(1 - a \cos x) + b \sin x}{x^3} = \frac{1}{3} \).

Steps to Solve

1. Expand the Functions To solve this limit, we need to expand ( 1 - a \cos x ) and ( b \sin x ) using Taylor series around ( x = 0 ):

• The expansion of ( \cos x ) is ( 1 - \frac{x^2}{2} + O(x^4) )
• The expansion of ( \sin x ) is ( x - \frac{x^3}{6} + O(x^5) )

Thus, we have: $$ 1 - a \cos x \approx 1 - a \left(1 - \frac{x^2}{2}\right) = 1 - a + \frac{ax^2}{2} $$

And: $$ b \sin x \approx b \left(x - \frac{x^3}{6}\right) = bx - \frac{bx^3}{6} $$

2. Combine the Expanded Terms Now, substituting back into the limit expression, we can combine: $$ x(1 - a \cos x) + b \sin x \approx x(1 - a + \frac{ax^2}{2}) + bx - \frac{bx^3}{6} $$

This becomes: $$ (1 - a + b)x + \frac{ax^3}{2} - \frac{bx^3}{6} $$

3. Form the Expression for the Limit Now we formulate the limit as: $$ \lim_{x \to 0} \frac{(1 - a + b)x + \left(\frac{a}{2} - \frac{b}{6}\right)x^3}{x^3} $$

Separating the terms gives us: $$ \lim_{x \to 0} \left(\frac{(1 - a + b)}{x^2} + \left(\frac{a}{2} - \frac{b}{6}\right)\right) $$

4. Setting the Conditions for the Limit to Exist For the limit to exist, the term involving ( x^2 ) must vanish, so we set: $$ 1 - a + b = 0 \implies b = a - 1 $$

5. Solving for ( a ) and ( b ) We need the constant term to equal ( \frac{1}{3} ): $$ \frac{a}{2} - \frac{(a - 1)}{6} = \frac{1}{3} $$

Converting and combining gives: $$ \frac{3a - (a - 1)}{6} = \frac{1}{3} $$ $$ \frac{2a + 1}{6} = \frac{1}{3} $$ $$ 2a + 1 = 2 \implies 2a = 1 \implies a = \frac{1}{2} $$

Substituting back gives: $$ b = \frac{1}{2} - 1 = -\frac{1}{2} $$