`lim_(x to 0) [(sin (x+a) + sin(a-x) -2sina)/(x sin x)]`=

Options

A. sin a
B. cos a
C. `-sin a`
D. `1/2 cos a`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin(x + a) + \sin(a - x) - 2\sin a}{x \sin x} \), we can follow these steps:

### Step 1: Simplify the numerator using the sine addition formula
We know that:
\[ \sin(x + a) = \sin x \cos a + \cos x \sin a \]
\[ \sin(a - x) = \sin a \cos x - \cos a \sin x \]
Substituting these into the limit gives:
\[ \sin(x + a) + \sin(a - x) = (\sin x \cos a + \cos x \sin a) + (\sin a \cos x - \cos a \sin x) \]
This simplifies to:
\[ (\sin x \cos a - \cos a \sin x) + 2 \cos x \sin a = 2 \cos x \sin a \]
Thus, the numerator becomes:
\[ 2 \cos x \sin a - 2 \sin a = 2 \sin a (\cos x - 1) \]

### Step 2: Substitute back into the limit
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{2 \sin a (\cos x - 1)}{x \sin x} \]

### Step 3: Analyze the limit
As \( x \to 0 \), both the numerator \( \cos x - 1 \) and the denominator \( x \sin x \) approach 0, leading to the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Taking the derivative of the numerator and the denominator:
- The derivative of the numerator \( 2 \sin a (\cos x - 1) \) is:
\[ -2 \sin a \sin x \]
- The derivative of the denominator \( x \sin x \) is:
\[ \sin x + x \cos x \]
Thus, we have:
\[ \lim_{x \to 0} \frac{-2 \sin a \sin x}{\sin x + x \cos x} \]

### Step 5: Substitute \( x = 0 \)
Now substituting \( x = 0 \):
- The numerator becomes \( -2 \sin a \cdot 0 = 0 \).
- The denominator becomes \( 0 + 0 \cdot 1 = 0 \), leading again to \( \frac{0}{0} \).

### Step 6: Apply L'Hôpital's Rule again
Taking the derivative again:
- The derivative of the numerator \( -2 \sin a \sin x \) is:
\[ -2 \sin a \cos x \]
- The derivative of the denominator \( \sin x + x \cos x \) is:
\[ \cos x + \cos x - x \sin x = 2 \cos x - x \sin x \]
Thus, we have:
\[ \lim_{x \to 0} \frac{-2 \sin a \cos x}{2 \cos x - x \sin x} \]

### Step 7: Substitute \( x = 0 \) again
Substituting \( x = 0 \):
- The numerator becomes \( -2 \sin a \cdot 1 = -2 \sin a \).
- The denominator becomes \( 2 \cdot 1 - 0 \cdot 0 = 2 \).

Thus, the limit evaluates to:
\[ \frac{-2 \sin a}{2} = -\sin a \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{\sin(x + a) + \sin(a - x) - 2\sin a}{x \sin x} = -\sin a \]