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 can use Taylor series expansions for \(\cos x\) and \(\sin x\).

### Step 1: Expand \(\cos x\) and \(\sin x\)

The Taylor series expansions around \(x = 0\) are:

\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \]

\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \]

### Step 2: Substitute the expansions into the limit

Substituting these expansions into the limit expression, we have:

\[ 1 + a \cos x - b \sin x = 1 + a\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) - b\left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) \]

This simplifies to:

\[ 1 + a - \frac{ax^2}{2} + \frac{ax^4}{24} - bx + \frac{bx^3}{6} - \frac{bx^5}{120} \]

### Step 3: Multiply by \(x\) and simplify

Now, multiplying the entire expression by \(x\):

\[ x(1 + a - bx - \frac{ax^2}{2} + \frac{ax^4}{24} + \frac{bx^3}{6}) = x(1 + a) - bx^2 - \frac{ax^3}{2} + \frac{ax^5}{24} + \frac{bx^4}{6} \]

### Step 4: Divide by \(x^3\)

Now, we divide the entire expression by \(x^3\):

\[ \frac{x(1 + a) - bx^2 - \frac{ax^3}{2} + \frac{ax^5}{24} + \frac{bx^4}{6}}{x^3} = \frac{1 + a}{x^2} - \frac{b}{x} - \frac{a}{2} + \frac{ax^2}{24} + \frac{bx}{6} \]

### Step 5: Analyze the limit as \(x \to 0\)

For the limit to equal 1, the coefficients of \(x^{-2}\) and \(x^{-1}\) must be zero:

1. Coefficient of \(x^{-2}\): \(1 + a = 0 \implies a = -1\)
2. Coefficient of \(x^{-1}\): \(-b = 0 \implies b = 0\)

### Step 6: Solve for \(a\) and \(b\)

Substituting \(a = -1\) into the second equation gives us:

\[ -\frac{-1}{2} + \frac{0}{6} = 1 \implies \frac{1}{2} = 1 \text{ (not true)} \]

We need to go back and check the coefficients of \(x^0\) and \(x^1\) for the limit to equal 1.

### Step 7: Set up the equations

From the constant term:

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

From the coefficient of \(x^3\):

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

### Step 8: Solve the system of equations

From equation (1):

\[ b = a + 1 \]

Substituting into equation (2):

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

Multiplying through by 6 to eliminate the fractions:

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

Substituting back to find \(b\):

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

### Step 9: Find \(ab\)

Now, we calculate \(ab\):

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

Thus, the value of \(ab\) is

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