If the value of lim _{x rightarrow 0}(2-cos x | vedclass

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(C) Let $L = \lim _{x \rightarrow 0}(2-\cos x \sqrt{\cos 2 x})^{\frac{x+2}{x^{2}}}$.This is of the form $1^{\infty}$.Using the formula $\lim _{x \righ

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$3$

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(C) Let $L = \lim _{x ightarrow 0}(2-\cos x \sqrt{\cos 2 x})^{\frac{x+2}{x^{2}}}$.This is of the form $1^{\infty}$.Using the formula $\lim _{x ightarrow 0} f(x)^{g(x)} = e^{\lim _{x ightarrow 0} (f(x)-1)g(x)}$,we get:$L = e^{\lim _{x ightarrow 0} (2-\cos x \sqrt{\cos 2 x}-1) \left(\frac{x+2}{x^{2}}ight)}$$L = e^{\lim _{x ightarrow 0} (1-\cos x \sqrt{\cos 2 x}) \left(\frac{x+2}{x^{2}}ight)}$As $x ightarrow 0$,$\frac{x+2}{x^2} \approx \frac{2}{x^2}$.So,$L = e^{\lim _{x ightarrow 0} \frac{1-\cos x \sqrt{\cos 2 x}}{x^2} 	imes 2}$.Let $f(x) = 1-\cos x \sqrt{\cos 2 x}$. Using Taylor series expansion:$\cos x \approx 1 - \frac{x^2}{2}$ and $\sqrt{\cos 2 x} = (1 - 2x^2)^{1/2} \approx 1 - x^2$.$f(x) \approx 1 - (1 - \frac{x^2}{2})(1 - x^2) = 1 - (1 - x^2 - \frac{x^2}{2} + \frac{x^4}{2}) \approx \frac{3x^2}{2}$.Thus,$\lim _{x ightarrow 0} \frac{f(x)}{x^2} = \frac{3}{2}$.Therefore,$L = e^{\frac{3}{2} 	imes 2} = e^3$.Comparing with $e^a$,we get $a = 3$.

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

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