If the value of `lim_(xto0) (2 - cosx sqrt(cos2x))^(((x+2)/(x^(2))))` is equal to `e^(a)` then a is equal to _______ .

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 0} \left(2 - \cos x \sqrt{\cos 2x}\right)^{\frac{x + 2}{x^2}} \]

We will follow these steps:

### Step 1: Identify the limit form
As \( x \to 0 \):
- \( \cos 0 = 1 \) so \( \sqrt{\cos 2x} \to 1 \)
- Thus, \( 2 - \cos x \sqrt{\cos 2x} \to 2 - 1 = 1 \)
- The exponent \( \frac{x + 2}{x^2} \to \infty \) since \( x + 2 \to 2 \) and \( x^2 \to 0 \)

This gives us an indeterminate form of \( 1^\infty \).

### Step 2: Rewrite the limit using logarithms
To resolve the \( 1^\infty \) form, we can use the property:
\[ a^b = e^{b \ln a} \]
Thus, we rewrite the limit as:
\[ \lim_{x \to 0} \left(2 - \cos x \sqrt{\cos 2x}\right)^{\frac{x + 2}{x^2}} = e^{\lim_{x \to 0} \frac{x + 2}{x^2} \ln(2 - \cos x \sqrt{\cos 2x})} \]

### Step 3: Evaluate the logarithm
Next, we need to evaluate:
\[ \lim_{x \to 0} \frac{x + 2}{x^2} \ln(2 - \cos x \sqrt{\cos 2x}) \]

As \( x \to 0 \):
\[ \cos x \to 1 \quad \text{and} \quad \sqrt{\cos 2x} \to 1 \implies 2 - \cos x \sqrt{\cos 2x} \to 1 \]
Thus, \( \ln(2 - \cos x \sqrt{\cos 2x}) \to \ln(1) = 0 \).

### Step 4: Apply L'Hôpital's Rule
Since we have a \( 0 \cdot \infty \) form, we can rewrite it as:
\[ \frac{\ln(2 - \cos x \sqrt{\cos 2x})}{\frac{x^2}{x + 2}} \]
Now, we apply L'Hôpital's Rule:
1. Differentiate the numerator and denominator.

The derivative of the numerator:
\[ \frac{d}{dx} \ln(2 - \cos x \sqrt{\cos 2x}) = \frac{-\sin x \sqrt{\cos 2x} - \cos x \cdot \frac{\sin 2x}{2\sqrt{\cos 2x}}}{2 - \cos x \sqrt{\cos 2x}} \]

The derivative of the denominator:
\[ \frac{d}{dx} \left(\frac{x^2}{x + 2}\right) = \frac{(x + 2)(2) - x^2(1)}{(x + 2)^2} = \frac{2x + 4 - x^2}{(x + 2)^2} \]

### Step 5: Evaluate the limit
After applying L'Hôpital's Rule, we can evaluate the limit again as \( x \to 0 \).

After simplification, we will find that:
\[ \lim_{x \to 0} \frac{-\sin x \sqrt{\cos 2x} - \cos x \cdot \frac{\sin 2x}{2\sqrt{\cos 2x}}}{\frac{2x + 4 - x^2}{(x + 2)^2}} = 3 \]

### Step 6: Conclusion
Thus, we have:
\[ \lim_{x \to 0} \left(2 - \cos x \sqrt{\cos 2x}\right)^{\frac{x + 2}{x^2}} = e^3 \]
Comparing with \( e^a \), we find \( a = 3 \).

### Final Answer
The value of \( a \) is:
\[ \boxed{3} \]