`f(x)` is the integral of `(2 sin x - sin 2x)/x^(3), x ne 0, lim_(x to 0) f'(x)`=………..

Answer

To solve the problem, we need to find the limit of \( f'(x) \) as \( x \) approaches 0, where \( f(x) \) is defined as the integral of \( \frac{2 \sin x - \sin 2x}{x^3} \) for \( x \neq 0 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \int \frac{2 \sin x - \sin 2x}{x^3} \, dx \]

2. **Differentiate \( f(x) \)** using the Fundamental Theorem of Calculus:
\[ f'(x) = \frac{2 \sin x - \sin 2x}{x^3} \]

3. **Evaluate the limit**:
\[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{x^3} \]
This limit results in the indeterminate form \( \frac{0}{0} \).

4. **Apply L'Hôpital's Rule**: Since we have an indeterminate form, we differentiate the numerator and the denominator:
- Derivative of the numerator \( 2 \sin x - \sin 2x \):
\[ \frac{d}{dx}(2 \sin x - \sin 2x) = 2 \cos x - 2 \cos 2x \]
- Derivative of the denominator \( x^3 \):
\[ \frac{d}{dx}(x^3) = 3x^2 \]

5. **Rewrite the limit**:
\[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2x}{3x^2} \]

6. **Evaluate the limit again**: This still results in the indeterminate form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:
- Derivative of the numerator \( 2 \cos x - 2 \cos 2x \):
\[ \frac{d}{dx}(2 \cos x - 2 \cos 2x) = -2 \sin x + 4 \sin 2x \]
- Derivative of the denominator \( 3x^2 \):
\[ \frac{d}{dx}(3x^2) = 6x \]

7. **Rewrite the limit again**:
\[ \lim_{x \to 0} \frac{-2 \sin x + 4 \sin 2x}{6x} \]

8. **Evaluate the limit**: This limit is still in the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule once more:
- Derivative of the numerator:
\[ \frac{d}{dx}(-2 \sin x + 4 \sin 2x) = -2 \cos x + 8 \cos 2x \]
- Derivative of the denominator:
\[ \frac{d}{dx}(6x) = 6 \]

9. **Final limit**:
\[ \lim_{x \to 0} \frac{-2 \cos x + 8 \cos 2x}{6} \]
Evaluating at \( x = 0 \):
\[ = \frac{-2 \cdot 1 + 8 \cdot 1}{6} = \frac{6}{6} = 1 \]

### Final Answer:
\[ \lim_{x \to 0} f'(x) = 1 \]