The value of `Lim_(x->0^+)(int_1^cosx(cos^-1t)dt)/(2x-sin(2x))`is equal to

Options

A. 0
B. `-1`
C. `2/3`
D. `(-1)/(4)`

Answer

To solve the limit

\[ \lim_{x \to 0^+} \frac{\int_1^{\cos x} \cos^{-1} t \, dt}{2x - \sin(2x)}, \]

we will follow these steps:

### Step 1: Identify the Form of the Limit
As \( x \to 0^+ \), both the numerator and the denominator approach 0. Thus, we have a \( \frac{0}{0} \) indeterminate form.

**Hint:** Check if direct substitution leads to an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

### Step 3: Differentiate the Numerator
Using Leibniz's rule for differentiation under the integral sign, we differentiate the numerator:

\[ \frac{d}{dx} \left( \int_1^{\cos x} \cos^{-1} t \, dt \right) = \cos^{-1}(\cos x) \cdot \frac{d}{dx}(\cos x) - 0 = \cos^{-1}(\cos x) \cdot (-\sin x). \]

Since \( \cos^{-1}(\cos x) = x \) for \( x \) in the range \( [0, \frac{\pi}{2}] \):

\[ \frac{d}{dx} \left( \int_1^{\cos x} \cos^{-1} t \, dt \right) = -x \sin x. \]

### Step 4: Differentiate the Denominator
Now we differentiate the denominator:

\[ \frac{d}{dx}(2x - \sin(2x)) = 2 - 2\cos(2x). \]

### Step 5: Rewrite the Limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0^+} \frac{-x \sin x}{2 - 2\cos(2x)}. \]

### Step 6: Substitute \( x = 0 \)
Substituting \( x = 0 \):

- The numerator becomes \( -0 \cdot \sin(0) = 0 \).
- The denominator becomes \( 2 - 2\cos(0) = 2 - 2 = 0 \).

We still have a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 7: Differentiate Again
Differentiate the numerator and denominator again:

1. **Numerator:**
\[ \frac{d}{dx}(-x \sin x) = -\sin x - x \cos x. \]

2. **Denominator:**
\[ \frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x). \]

### Step 8: Rewrite the Limit Again
Now we have:

\[ \lim_{x \to 0^+} \frac{-\sin x - x \cos x}{4\sin(2x)}. \]

### Step 9: Substitute \( x = 0 \) Again
Substituting \( x = 0 \):

- The numerator becomes \( -\sin(0) - 0 \cdot \cos(0) = 0 \).
- The denominator becomes \( 4\sin(0) = 0 \).

We still have a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule one more time.

### Step 10: Differentiate Again
Differentiate the numerator and denominator again:

1. **Numerator:**
\[ \frac{d}{dx}(-\sin x - x \cos x) = -\cos x - (\cos x - x \sin x) = -2\cos x + x \sin x. \]

2. **Denominator:**
\[ \frac{d}{dx}(4\sin(2x)) = 8\cos(2x). \]

### Step 11: Rewrite the Limit Again
Now we have:

\[ \lim_{x \to 0^+} \frac{-2\cos x + x \sin x}{8\cos(2x)}. \]

### Step 12: Substitute \( x = 0 \) Again
Substituting \( x = 0 \):

- The numerator becomes \( -2\cos(0) + 0 \cdot \sin(0) = -2 \).
- The denominator becomes \( 8\cos(0) = 8 \).

### Final Result
Thus, the limit evaluates to:

\[ \frac{-2}{8} = -\frac{1}{4}. \]

Therefore, the value of the limit is

\[ \boxed{-\frac{1}{4}}. \]