The value of `lim_(x rarr0)(2int_(0)^(cos x ) cos^(-1) (t))/(2x- sin 2 x)dx` is

Options

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

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{2 \int_0^{\cos x} \cos^{-1}(t) \, dt}{2x - \sin 2x}, \]

we will follow these steps:

### Step 1: Identify the Indeterminate Form
When we substitute \( x = 0 \) into the limit, we find that both the numerator and denominator approach 0. This gives us the indeterminate form \( \frac{0}{0} \).

**Hint:** Check the values of the numerator and denominator at \( x = 0 \) to confirm the indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

**Hint:** Remember that L'Hôpital's Rule is applicable for \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms.

### Step 3: Differentiate the Numerator
The numerator is \( 2 \int_0^{\cos x} \cos^{-1}(t) \, dt \). We need to differentiate this with respect to \( x \). By using Leibniz's rule for differentiation under the integral sign, we have:

\[ \frac{d}{dx} \left( \int_0^{\cos x} \cos^{-1}(t) \, dt \right) = \cos^{-1}(\cos x) \cdot \frac{d}{dx}(\cos x) - \cos^{-1}(0) \cdot 0. \]

Since \( \cos^{-1}(\cos x) = x \) and \( \frac{d}{dx}(\cos x) = -\sin x \), we get:

\[ \frac{d}{dx} \left( 2 \int_0^{\cos x} \cos^{-1}(t) \, dt \right) = 2 \left( x \cdot (-\sin x) \right) = -2x \sin x. \]

**Hint:** Use Leibniz's rule carefully, noting the limits and the derivatives involved.

### Step 4: Differentiate the Denominator
The denominator is \( 2x - \sin 2x \). Differentiating this gives:

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

**Hint:** Remember the derivative of \( \sin kx \) is \( k \cos kx \).

### Step 5: Set Up the New Limit
Now we can set up the limit again using the derivatives we found:

\[ \lim_{x \to 0} \frac{-2x \sin x}{2 - 2\cos 2x}. \]

**Hint:** Substitute \( x = 0 \) into the new limit to check if it still results in an indeterminate form.

### Step 6: Simplify the New Limit
As \( x \to 0 \), \( \sin x \to 0 \) and \( \cos 2x \to 1 \). Thus, the denominator approaches \( 2 - 2(1) = 0 \), leading to another \( \frac{0}{0} \) form.

### Step 7: Apply L'Hôpital's Rule Again
We apply L'Hôpital's Rule again:

1. Differentiate the numerator:
\(-2(\sin x + x \cos x)\).

2. Differentiate the denominator:
\(-4\sin 2x\).

### Step 8: Evaluate the New Limit
Now we have:

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

As \( x \to 0 \), \( \sin x \to x \) and \( \cos x \to 1 \), so the numerator approaches \( 0 + 0 = 0 \) and the denominator approaches \( 0 \).

### Step 9: Apply L'Hôpital's Rule Once More
Differentiate again:

1. Numerator: \( \cos x + \cos x - x \sin x \).
2. Denominator: \( 4\cos 2x \).

Evaluating this limit as \( x \to 0 \):

\[ \lim_{x \to 0} \frac{2\cos x}{4\cos 2x} = \frac{2(1)}{4(1)} = \frac{1}{2}. \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{-\frac{1}{2}}. \]