We need to evaluate

$$L=\lim_{x\to 0}\left(2-\cos x\,\sqrt{\cos 2x}\right)^{\frac{x+2}{x^2}}$$

and it is given that $L=e^a$. We must find $a$.

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1. Identify the indeterminate form

As $x\to 0$,

• $\cos x \to 1$
• $\cos 2x \to 1$, so $\sqrt{\cos 2x}\to 1$

Hence,

$$2-\cos x\sqrt{\cos 2x} \to 2-1=1$$

and

$$\frac{x+2}{x^2}\to \infty.$$

So the expression is of the form $1^{\infty}$.

We use the standard method: If

$$L=\lim (1+u(x))^{v(x)},$$

then

$$\ln L=\lim v(x)\ln(1+u(x)).$$

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2. Expand the base near $x=0$

We need the expansion of

$$2-\cos x\sqrt{\cos 2x}.$$

(i) Expansion of $\cos x$

$$\cos x = 1-\frac{x^2}{2}+O(x^4).$$

(ii) Expansion of $\cos 2x$

$$\cos 2x = 1-\frac{(2x)^2}{2}+O(x^4)=1-2x^2+O(x^4).$$

(iii) Expansion of $\sqrt{\cos 2x}$

Using

$$\sqrt{1+t}=1+\frac t2+O(t^2),$$

with $t=-2x^2+O(x^4)$,

$$\sqrt{\cos 2x}=1- x^2+O(x^4).$$

(iv) Product

Now,

$$\cos x\sqrt{\cos 2x} =\left(1-\frac{x^2}{2}+O(x^4)\right)\left(1-x^2+O(x^4)\right).$$

Multiplying up to $x^2$ terms,

$$\cos x\sqrt{\cos 2x}=1-\frac{x^2}{2}-x^2+O(x^4) =1-\frac{3x^2}{2}+O(x^4).$$

Therefore,

$$2-\cos x\sqrt{\cos 2x} =2-\left(1-\frac{3x^2}{2}+O(x^4)\right) =1+\frac{3x^2}{2}+O(x^4).$$

So the base is

$$1+\frac{3x^2}{2}+O(x^4).$$

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3. Take logarithm

Let

$$L=\lim_{x\to 0}\left(2-\cos x\sqrt{\cos 2x}\right)^{\frac{x+2}{x^2}}.$$

Then

$$\ln L=\lim_{x\to 0}\frac{x+2}{x^2}\ln\left(2-\cos x\sqrt{\cos 2x}\right).$$

Using

$$2-\cos x\sqrt{\cos 2x}=1+\frac{3x^2}{2}+O(x^4),$$

we get

$$\ln\left(2-\cos x\sqrt{\cos 2x}\right) =\ln\left(1+\frac{3x^2}{2}+O(x^4)\right) =\frac{3x^2}{2}+O(x^4).$$

Therefore,

$$\ln L=\lim_{x\to 0}\frac{x+2}{x^2}\left(\frac{3x^2}{2}+O(x^4)\right).$$

Simplify:

$$\ln L=\lim_{x\to 0}(x+2)\left(\frac{3}{2}+O(x^2)\right).$$

Now let $x\to 0$:

$$\ln L=2\cdot \frac{3}{2}=3.$$

Hence,

$$L=e^3.$$

So if $L=e^a$, then

$$a=3.$$

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4. Final answer

$$\boxed{3}$$

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5. Comparison with stored correct answer

Stored correct answer: $3$

Our derived answer matches the stored correct answer.