The value of ordered pair (a,b) such that `lim _(xto0) (x (1+ a cos x ) -b sin x )/( x ^(3))=1,` is:

Options

A. `(-(5)/(2), -(3)/(2))`
B. `((5)/(2), (3)/(2))`
C. `(-(5)/(2),( 3)/(2))`
D. `((5 )/(2),(3)/(2))`

Answer

To solve the limit problem, we need to find the ordered pair \((a, b)\) such that:

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

### Step-by-Step Solution:

1. **Rewrite the Limit**:
We start by rewriting the limit expression:
\[ \lim_{x \to 0} \frac{x(1 + a \cos x - b \sin x)}{x^3} = \lim_{x \to 0} \frac{1 + a \cos x - b \sin x}{x^2} \]

2. **Expand \(\cos x\) and \(\sin x\)**:
Using Taylor series expansions around \(x = 0\):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]

3. **Substitute the Expansions**:
Substitute these expansions into the limit:
\[ 1 + a(1 - \frac{x^2}{2}) - b(x - \frac{x^3}{6}) = 1 + a - \frac{a}{2}x^2 - bx + \frac{b}{6}x^3 + O(x^4) \]

4. **Combine Like Terms**:
Rearranging gives:
\[ (1 + a) - bx - \frac{a}{2}x^2 + \frac{b}{6}x^3 + O(x^4) \]

5. **Set Up the Limit**:
Now, we can write the limit:
\[ \lim_{x \to 0} \frac{(1 + a) - bx - \frac{a}{2}x^2 + \frac{b}{6}x^3 + O(x^4)}{x^2} \]

6. **Analyze the Limit**:
For the limit to exist and equal 1, we need:
- The constant term \(1 + a\) must be \(0\) (to avoid divergence).
- The coefficient of \(x^2\) must equal \(-1\) (to ensure the limit equals 1).

7. **Set Up the Equations**:
From \(1 + a = 0\):
\[ a = -1 \]
From \(-\frac{a}{2} + \frac{b}{6} = 1\):
\[ -\frac{-1}{2} + \frac{b}{6} = 1 \implies \frac{1}{2} + \frac{b}{6} = 1 \]
\[ \frac{b}{6} = 1 - \frac{1}{2} = \frac{1}{2} \implies b = 3 \]

8. **Final Ordered Pair**:
Thus, the ordered pair \((a, b)\) is:
\[ (a, b) = (-1, 3) \]

### Final Answer:
The value of the ordered pair \((a, b)\) is \((-1, 3)\).