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 find the ordered pair \((a, b)\) such that

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

we can follow these steps:

### Step 1: Analyze the limit
The expression inside the limit is of the form \(\frac{0}{0}\) as \(x \to 0\). Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator:

1. **Numerator**:
\[ \frac{d}{dx}[x(1 + a \cos x) - b \sin x] \]
Using the product rule for \(x(1 + a \cos x)\):
\[ = (1 + a \cos x) + x(-a \sin x) - b \cos x \]
Simplifying gives:
\[ = 1 + a \cos x - ax \sin x - b \cos x = 1 + (a - b) \cos x - ax \sin x \]

2. **Denominator**:
\[ \frac{d}{dx}[x^3] = 3x^2 \]

### Step 3: Rewrite the limit
Now we have:
\[ \lim_{x \to 0} \frac{1 + (a - b) \cos x - ax \sin x}{3x^2} \]

### Step 4: Evaluate the limit again
As \(x \to 0\), this limit is still of the form \(\frac{0}{0}\). We apply L'Hôpital's Rule again.

1. **Numerator**:
\[ \frac{d}{dx}[1 + (a - b) \cos x - ax \sin x] = -(a - b) \sin x - (a \sin x + ax \cos x) \]
Simplifying gives:
\[ = -(a - b) \sin x - a \sin x - ax \cos x = -(a + (a - b)) \sin x - ax \cos x \]

2. **Denominator**:
\[ \frac{d}{dx}[3x^2] = 6x \]

### Step 5: Rewrite the limit again
Now we have:
\[ \lim_{x \to 0} \frac{-(a + (a - b)) \sin x - ax \cos x}{6x} \]

### Step 6: Evaluate the limit as \(x \to 0\)
As \(x \to 0\), \(\sin x \to x\) and \(\cos x \to 1\):
\[ \lim_{x \to 0} \frac{-(a + (a - b))x - ax}{6x} = \frac{-(2a - b)}{6} \]

Setting this equal to 1 gives:
\[ \frac{-(2a - b)}{6} = 1 \implies -(2a - b) = 6 \implies 2a - b = -6 \quad \text{(Equation 1)} \]

### Step 7: Set up the first equation
From the original limit, we also have:
\[ 1 + (a - b) - b = 0 \implies a - 2b + 1 = 0 \quad \text{(Equation 2)} \]

### Step 8: Solve the equations
We have two equations:
1. \(2a - b = -6\)
2. \(a - 2b + 1 = 0\)

From Equation 2, we can express \(a\) in terms of \(b\):
\[ a = 2b - 1 \]

Substituting this into Equation 1:
\[ 2(2b - 1) - b = -6 \implies 4b - 2 - b = -6 \implies 3b - 2 = -6 \implies 3b = -4 \implies b = -\frac{4}{3} \]

Now substituting \(b\) back to find \(a\):
\[ a = 2(-\frac{4}{3}) - 1 = -\frac{8}{3} - 1 = -\frac{8}{3} - \frac{3}{3} = -\frac{11}{3} \]

### Final Result
Thus, the ordered pair \((a, b)\) is:
\[ \left(-\frac{11}{3}, -\frac{4}{3}\right) \]