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Step-by-Step Solution
Step 1: Recognize the Limit Form
We are asked to evaluate the limit
\displaystyle \lim_{x \to 0} \left(2 - \cos x \,\sqrt{\cos 2x}\right)^{\frac{x + 2}{x^2}}.
As x \to 0 , we have
\cos 0 = 1 and \sqrt{\cos(2 \cdot 0)} = \sqrt{1} = 1.
Thus,
2 - \cos 0 \,\sqrt{\cos 2 \cdot 0} = 2 - 1 = 1.
So the base of the power approaches 1, while \frac{x+2}{x^2} becomes very large (tends to infinity). This is an indeterminate form of the type 1^\infty.
Step 2: Transform the Expression into the Exponential Form
A standard technique for limits of the type 1^\infty is to write the expression in the form
\bigl(1 + f(x)\bigr)^{g(x)} \;=\; e^{\,\lim\limits_{x \to 0} \left[f(x)\,g(x)\right]},
where f(x) = \left(2 - \cos x \,\sqrt{\cos 2x}\right) - 1 and g(x) = \frac{x + 2}{x^2}.
Hence we attempt to compute
\displaystyle \lim_{x \to 0} \left(2 - \cos x \,\sqrt{\cos 2x}\right)^{\frac{x + 2}{x^2}}
= e^{\,\lim\limits_{x \to 0} \left[\bigl(\,2 - \cos x \,\sqrt{\cos 2x} - 1\bigr)\,\cdot \frac{x+2}{x^2}\right]}.
Notice that
2 - \cos x \,\sqrt{\cos 2x} - 1 = 1 - \cos x \,\sqrt{\cos 2x}.
Step 3: Focus on the Key Inner Limit
Define
\displaystyle L = \lim_{x \to 0} \frac{\,1 - \cos x \,\sqrt{\cos 2x}\,}{x^2}.
Once we find L, we then multiply it by (x+2) in the exponent and see what happens as x \to 0.
So we want to evaluate
\displaystyle \lim_{x \to 0} \left[\frac{\,1 - \cos x \,\sqrt{\cos 2x}\,}{x^2}\right] \cdot (x+2).
Step 4: Compute \displaystyle \lim_{x \to 0} \frac{1 - \cos x \,\sqrt{\cos 2x}}{x^2}
This can be tricky to compute directly. One possible approach is to expand using series or apply L'Hospital’s Rule. Let us outline a simpler series expansion approach (although L'Hospital’s can also be used):
As x \to 0, \cos x \approx 1 - \frac{x^2}{2}, \quad \cos(2x) \approx 1 - 2x^2.
Hence, \sqrt{\cos 2x} \approx \sqrt{\,1 - 2x^2\,} \approx 1 - x^2 \quad (to first non-trivial terms).
So
\cos x \,\sqrt{\cos 2x}
\approx \bigl(\,1 - \tfrac{x^2}{2}\bigr)\,\bigl(\,1 - x^2\bigr)
= 1 - \tfrac{x^2}{2} - x^2 + O\bigl(x^4\bigr)
= 1 - \tfrac{3x^2}{2} + O\bigl(x^4\bigr).
Therefore,
1 - \cos x \,\sqrt{\cos 2x} \approx 1 - \bigl(1 - \tfrac{3x^2}{2}\bigr)
= \tfrac{3x^2}{2}.
Dividing by x^2, we get
\frac{1 - \cos x \,\sqrt{\cos 2x}}{x^2} \approx \frac{\tfrac{3x^2}{2}}{x^2} = \tfrac{3}{2}.
Thus,
L = \lim_{x \to 0} \frac{\,1 - \cos x \,\sqrt{\cos 2x}\,}{x^2} = \tfrac{3}{2}.
Step 5: Multiply by (x+2) and Find the Overall Exponent
The exponent to which e is raised is then
\displaystyle \lim_{x \to 0} \left[\frac{1 - \cos x \,\sqrt{\cos 2x}}{x^2} \cdot (x+2)\right]
= \left(\frac{3}{2}\right)\,\lim_{x \to 0} (x+2).
Since \lim\limits_{x \to 0} (x + 2) = 2, we get
\left(\frac{3}{2}\right) \times 2 = 3.
Step 6: Conclude the Value of the Limit
Hence the original limit is
\displaystyle e^{\,3}.
In the problem statement, the expression is given to be e^a. Therefore, we equate
e^a = e^3 \quad\Longrightarrow\quad a = 3.
Final Answer
a = 3
