The value of `lim _( x to 0^(+))(int _(1)^(cos x) (cos ^(-1) t )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: Evaluate the limit directly
First, we substitute \(x = 0\) into the limit:

\[ \int_{1}^{\cos(0)} \cos^{-1} t \, dt = \int_{1}^{1} \cos^{-1} t \, dt = 0, \]
and
\[ 2(0) - \sin(2 \cdot 0) = 0 - 0 = 0. \]

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator separately.

#### Numerator:
Using the Fundamental Theorem of Calculus, we differentiate the numerator:

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

#### Denominator:
Now we differentiate the denominator:

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

### Step 3: Rewrite the limit
Now we rewrite the limit using the derivatives we found:

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

### Step 4: Substitute \(x = 0\) again
Substituting \(x = 0\) again gives us:

\[ \text{Numerator: } -0 \cdot \sin(0) = 0, \]
and
\[ \text{Denominator: } 2 - 2\cos(0) = 2 - 2 = 0. \]

Again, we have the indeterminate form \( \frac{0}{0} \).

### Step 5: Apply L'Hôpital's Rule again
We apply L'Hôpital's Rule again.

#### New Numerator:
Differentiating the numerator again gives:

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

#### New Denominator:
Differentiating the denominator gives:

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

### Step 6: Rewrite the limit again
Now we have:

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

### Step 7: Substitute \(x = 0\) once more
Substituting \(x = 0\):

\[ \text{Numerator: } -\sin(0) - 0 \cdot \cos(0) = 0, \]
and
\[ \text{Denominator: } 4 \sin(0) = 0. \]

We still have \( \frac{0}{0} \).

### Step 8: Apply L'Hôpital's Rule a third time
We apply L'Hôpital's Rule again.

#### New Numerator:
Differentiating gives:

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

#### New Denominator:
Differentiating gives:

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

### Step 9: Rewrite the limit again
Now we have:

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

### Step 10: Substitute \(x = 0\) one last time
Substituting \(x = 0\):

\[ \text{Numerator: } -2\cos(0) + 0 \cdot \sin(0) = -2, \]
and
\[ \text{Denominator: } 8\cos(0) = 8. \]

### Final Result
Thus, we find:

\[ \lim_{x \to 0^{+}} \frac{-2}{8} = -\frac{1}{4}. \]

So, the value of the limit is

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