If `lim_(x to 0) (x(1 + a cos x)-b sinx)/(x^3) = 1`, then the value of ab is euqal to

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{x(1 + a \cos x) - b \sin x}{x^3} = 1, \]

we will use Taylor series expansions for \(\cos x\) and \(\sin x\) around \(x = 0\).

### Step 1: Write the Taylor series expansions
The Taylor series expansions around \(x = 0\) are:
- \(\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)\)
- \(\sin x \approx x - \frac{x^3}{6} + O(x^5)\)

### Step 2: Substitute the expansions into the limit
Substituting these expansions into the limit expression, we have:

\[ \cos x \approx 1 - \frac{x^2}{2}, \quad \sin x \approx x - \frac{x^3}{6}. \]

Thus,

\[ 1 + a \cos x \approx 1 + a \left(1 - \frac{x^2}{2}\right) = 1 + a - \frac{a x^2}{2}. \]

Now substituting into the original expression:

\[ x(1 + a \cos x) - b \sin x \approx x\left(1 + a - \frac{a x^2}{2}\right) - b\left(x - \frac{x^3}{6}\right). \]

### Step 3: Simplify the expression
Expanding this gives:

\[ x(1 + a) - \frac{a x^3}{2} - bx + \frac{b x^3}{6}. \]

Combining like terms, we get:

\[ (1 + a - b)x + \left(-\frac{a}{2} + \frac{b}{6}\right)x^3. \]

### Step 4: Divide by \(x^3\) and take the limit
Now we divide the entire expression by \(x^3\):

\[ \frac{(1 + a - b)x + \left(-\frac{a}{2} + \frac{b}{6}\right)x^3}{x^3} = \frac{(1 + a - b)}{x^2} + \left(-\frac{a}{2} + \frac{b}{6}\right). \]

As \(x \to 0\), the term \(\frac{(1 + a - b)}{x^2}\) will approach infinity unless \(1 + a - b = 0\). Therefore, we must have:

\[ 1 + a - b = 0 \quad \text{(Equation 1)}. \]

### Step 5: Set the remaining term equal to 1
For the limit to equal 1, we need:

\[ -\frac{a}{2} + \frac{b}{6} = 1 \quad \text{(Equation 2)}. \]

### Step 6: Solve the equations
From Equation 1, we can express \(b\) in terms of \(a\):

\[ b = 1 + a. \]

Substituting this into Equation 2:

\[ -\frac{a}{2} + \frac{1 + a}{6} = 1. \]

Multiplying through by 6 to eliminate the fractions:

\[ -3a + 1 + a = 6 \implies -2a + 1 = 6 \implies -2a = 5 \implies a = -\frac{5}{2}. \]

Now substituting \(a\) back into Equation 1 to find \(b\):

\[ b = 1 - \frac{5}{2} = -\frac{3}{2}. \]

### Step 7: Calculate \(ab\)
Now we can find \(ab\):

\[ ab = \left(-\frac{5}{2}\right) \left(-\frac{3}{2}\right) = \frac{15}{4}. \]

Thus, the value of \(ab\) is

\[ \boxed{\frac{15}{4}}. \]