`L=lim_(xrarr0) (sin(sinx)-sinx)/(ax^(5)+bx^(3)+c)=-(1)/(12)`
The value/values of a is

Options

A. `in` R
B. 2
C. 0
D. 1

Explanation

To solve the limit problem given by

\[ L = \lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{ax^5 + bx^3 + c} = -\frac{1}{12} \]

we will follow these steps:

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0. Therefore, we have a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule.

### Step 2: Set up the limit
The limit can be rewritten as:

\[ L = \lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{ax^5 + bx^3 + c} \]

### Step 3: Determine the condition for \( c \)
For the limit to exist and not be equal to 0, we need to ensure that the denominator does not approach a non-zero value when the numerator is 0. Thus, we set \( c = 0 \) to ensure the denominator approaches 0 as well.

### Step 4: Substitute \( c = 0 \)
Now the limit simplifies to:

\[ L = \lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{ax^5 + bx^3} \]

### Step 5: Use Taylor series expansion
Using the Taylor series expansion for small \( x \):

- \( \sin x \approx x - \frac{x^3}{6} + O(x^5) \)
- Therefore, \( \sin(\sin x) \approx \sin\left(x - \frac{x^3}{6}\right) \)

Using the Taylor expansion for \( \sin \):

\[ \sin\left(x - \frac{x^3}{6}\right) \approx \left(x - \frac{x^3}{6}\right) - \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3 + O(x^5) \]

Calculating \( \sin(\sin x) - \sin x \):

\[ \sin(\sin x) - \sin x \approx \left(-\frac{x^3}{6} + O(x^5)\right) - \left(-\frac{x^3}{6} + O(x^5)\right) = O(x^5) \]

### Step 6: Substitute back into the limit
Now we have:

\[ L = \lim_{x \to 0} \frac{O(x^5)}{ax^5 + bx^3} \]

### Step 7: Factor out \( x^3 \) from the denominator
Factoring out \( x^3 \):

\[ L = \lim_{x \to 0} \frac{O(x^5)}{x^3(ax^2 + b)} = \lim_{x \to 0} \frac{O(x^5)}{x^3} \cdot \frac{1}{ax^2 + b} \]

### Step 8: Evaluate the limit
As \( x \to 0 \), \( O(x^5)/x^3 \to 0 \) and \( ax^2 + b \to b \). Thus, the limit becomes:

\[ L = \lim_{x \to 0} \frac{0}{b} = 0 \]

### Step 9: Apply L'Hôpital's Rule
Since we have \( 0/0 \) form, we apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

1. For the numerator: \( \frac{d}{dx}(\sin(\sin x) - \sin x) = \cos(\sin x) \cos x - \cos x \)
2. For the denominator: \( \frac{d}{dx}(ax^5 + bx^3) = 5ax^4 + 3bx^2 \)

### Step 10: Evaluate the new limit
Substituting \( x = 0 \):

\[ L = \frac{\cos(0)(\cos(0) - 1)}{0} = \frac{1(1 - 1)}{0} = 0 \]

### Step 11: Set up the equation
We need to find \( a \) such that:

\[ L = -\frac{1}{12} \]

From the previous steps, we can conclude that:

\[ \frac{-1}{6b} = -\frac{1}{12} \implies 6b = 12 \implies b = 2 \]

### Conclusion
Since \( b = 2 \) and \( c = 0 \), \( a \) can take any real value. Thus, the value of \( a \) is:

\[ \text{The value of } a \text{ is any real number.} \]