`lim_(xrarr -1) (cos 2 -cos 2x)/(x^2-|x|)` is equaol to

Options

A. 2
B. `sin 2`
C. `2sin2`
D. none of these

Explanation

To solve the limit \( \lim_{x \to -1} \frac{\cos 2 - \cos 2x}{x^2 - |x|} \), we can follow these steps:

### Step 1: Rewrite the limit expression
We start with the limit:
\[ \lim_{x \to -1} \frac{\cos 2 - \cos 2x}{x^2 - |x|} \]
Since \( x \) is approaching \(-1\), we note that \( |x| = -x \) when \( x < 0 \). Thus, we can rewrite the denominator:
\[ x^2 - |x| = x^2 + x \]
So the limit becomes:
\[ \lim_{x \to -1} \frac{\cos 2 - \cos 2x}{x^2 + x} \]

### Step 2: Use the cosine difference identity
We can use the identity for the difference of cosines:
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
In our case, let \( A = 2 \) and \( B = 2x \):
\[ \cos 2 - \cos 2x = -2 \sin\left(\frac{2 + 2x}{2}\right) \sin\left(\frac{2 - 2x}{2}\right) = -2 \sin(1 + x) \sin(1 - x) \]
Thus, the limit now looks like:
\[ \lim_{x \to -1} \frac{-2 \sin(1 + x) \sin(1 - x)}{x^2 + x} \]

### Step 3: Substitute the denominator
Now substitute \( x^2 + x \):
\[ x^2 + x = x(x + 1) \]
So we have:
\[ \lim_{x \to -1} \frac{-2 \sin(1 + x) \sin(1 - x)}{x(x + 1)} \]

### Step 4: Evaluate the limit
As \( x \to -1 \):
- \( 1 + x \to 0 \)
- \( 1 - x \to 2 \)

Thus, we can evaluate the limit:
\[ \lim_{x \to -1} \frac{-2 \sin(1 + x) \sin(1 - x)}{x(x + 1)} = \lim_{x \to -1} \frac{-2 \sin(0) \sin(2)}{x(x + 1)} \]
Since \( \sin(0) = 0 \), the numerator approaches \( 0 \).

### Step 5: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- Derivative of the numerator:
\[ \frac{d}{dx}[-2 \sin(1 + x) \sin(1 - x)] = -2 \left( \cos(1 + x) \sin(1 - x) - \sin(1 + x) \cos(1 - x) \right) \]
- Derivative of the denominator:
\[ \frac{d}{dx}[x(x + 1)] = 2x + 1 \]

### Step 6: Evaluate the derivatives at \( x = -1 \)
Now we evaluate:
\[ \lim_{x \to -1} \frac{-2 \left( \cos(1 - 1) \sin(1 + 1) - \sin(1 - 1) \cos(1 + 1) \right)}{2(-1) + 1} \]
This simplifies to:
\[ \lim_{x \to -1} \frac{-2 \left( \cos(0) \sin(2) - 0 \cdot \cos(2) \right)}{-1} = \frac{-2(1)(\sin(2))}{-1} = 2 \sin(2) \]

### Final Result
Thus, the limit is:
\[ \boxed{2 \sin(2)} \]