If `lim_(x to 0)(x(1+ a cos x)- b sin x)/x^(3)=1`, then a,b are

Options

A. `1/2, -3/2`
B. `5/2, 3/2`
C. `-5/2, -3/2`
D. none

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 first analyze the expression inside the limit and use Taylor series expansions for \(\cos x\) and \(\sin x\).

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

Using Taylor series expansion around \(x = 0\):

\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6), \]
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7). \]

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

Substituting these expansions into the limit expression, we get:

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

This simplifies to:

\[ 1 + a - b x - \frac{a x^2}{2} + \frac{b x^3}{6} + O(x^4). \]

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

Now, we multiply the entire expression by \(x\):

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

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

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

\[ \frac{x(1 + a) - b x^2 - \frac{a x^3}{2} + O(x^4)}{x^3} = \frac{(1 + a)}{x^2} - \frac{b}{x} - \frac{a}{2} + O(1). \]

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

As \(x \to 0\), the terms \(\frac{(1 + a)}{x^2}\) and \(-\frac{b}{x}\) will dominate unless they are set to zero. For the limit to equal 1, we need:

1. The coefficient of \(\frac{1}{x^2}\) to be zero:
\[ 1 + a = 0 \implies a = -1. \]

2. The coefficient of \(\frac{1}{x}\) to be zero:
\[ -b = 0 \implies b = 0. \]

### Step 6: Substitute values back into the limit

Now substituting \(a = -1\) and \(b = 0\) back into the limit expression:

\[ \lim_{x \to 0} \left(-\frac{1}{2} + O(1)\right) = 1. \]

### Conclusion

Thus, the values of \(a\) and \(b\) are:

\[ \boxed{(-1, 0)}. \]