If the value of $$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{\left( {{{x + 2} \over {{x^2}}}} \right)}}$$ is equal to e^a, then a is equal to __________.

$$\mathop {\lim }\limits_{x \to 0} {(2 - \cos x\sqrt {\cos 2x} )^{{{x + 2} \over {{x^2}}}}}$$<br><br>form : 1^$\infty$ $$ = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right) \times (x + 2)}}$$<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}} = \mathop {\lim }\limits_{x \to 0} {{\sin x\sqrt {\cos 2x} - \cos x \times {1 \over {2\sqrt {\cos 2x} }} \times ( - 2sin2x)} \over {2x}}$$ (by L' Hospital Rule)<br><br>$$\mathop {\lim }\limits_{x \to 0} {{\sin x\cos 2x + \sin 2x.\cos x} \over {2x}} = {1 \over 2} + 1 = {3 \over 2}$$<br><br>So, $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 - \cos x\sqrt {\cos 2x} } \over {{x^2}}}} \right)(x + 2)}}$$<br><br>$= {e^{{3 \over 2} \times 2}} = {e^3}$<br><br>$\Rightarrow$ a = 3