If f(x) is the integral of `(2 sin x-sin 2 x)/(x^(3)), "where x" ne 0, "then find" underset(x rarr 0)("lim")f'(x)`.

Answer

To solve the problem, we need to find the limit as \( x \) approaches 0 of the derivative of the function \( f(x) \), which is defined as the integral of \( \frac{2 \sin x - \sin 2x}{x^3} \). Let's break down the solution step by step.

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is given by:
\[ f(x) = \int \frac{2 \sin x - \sin 2x}{x^3} \, dx \]

### Step 2: Differentiate \( f(x) \) to find \( f'(x) \)
Using the Fundamental Theorem of Calculus, we can differentiate \( f(x) \):
\[ f'(x) = \frac{2 \sin x - \sin 2x}{x^3} \]

### Step 3: Substitute \( \sin 2x \) using the double angle formula
Recall that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite \( f'(x) \):
\[ f'(x) = \frac{2 \sin x - 2 \sin x \cos x}{x^3} = \frac{2 \sin x (1 - \cos x)}{x^3} \]

### Step 4: Find the limit as \( x \) approaches 0
We need to evaluate:
\[ \lim_{x \to 0} f'(x) = \lim_{x \to 0} \frac{2 \sin x (1 - \cos x)}{x^3} \]

### Step 5: Break the limit into parts
We can separate the limit:
\[ \lim_{x \to 0} f'(x) = 2 \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{(1 - \cos x)}{x^2} \]

### Step 6: Evaluate the limits
1. We know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
2. For the second limit, we use the fact that:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
(This can be derived from the Taylor series expansion of \( \cos x \) or using L'Hôpital's rule.)

### Step 7: Combine the results
Now substituting back into our limit:
\[ \lim_{x \to 0} f'(x) = 2 \cdot 1 \cdot \frac{1}{2} = 1 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} f'(x) = 1 \]