mathop {Lim}limits_{x to 0} frac{{log _{{{sin | vedclass

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(C) Let $L = \mathop {Lim}\limits_{x 	o 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$.Using the base chan

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$4$

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(C) Let $L = \mathop {Lim}\limits_{x 	o 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$.Using the base change formula $\log_a b = \frac{\ln b}{\ln a}$,we get:$L = \mathop {Lim}\limits_{x 	o 0} \frac{\ln(\cos x) / \ln(\sin^2 x)}{\ln(\cos(x/2)) / \ln(\sin^2(x/2))}$.As $x 	o 0$,$\cos x \approx 1 - x^2/2$ and $\sin x \approx x$,so $\ln(\cos x) \approx -x^2/2$ and $\ln(\sin^2 x) \approx \ln(x^2) = 2\ln x$.Similarly,$\ln(\cos(x/2)) \approx -(x/2)^2/2 = -x^2/8$ and $\ln(\sin^2(x/2)) \approx \ln(x^2/4) = 2\ln(x/2) = 2(\ln x - \ln 2) \approx 2\ln x$.Substituting these approximations:$L = \mathop {Lim}\limits_{x 	o 0} \frac{-x^2/2}{2\ln x} \cdot \frac{2\ln x}{-x^2/8} = \frac{-x^2/2}{-x^2/8} = \frac{8}{2} = 4$.

$\mathop {Lim}\limits_{x \to 0} \frac{{\log _{{{\sin }^2}x}}\cos x}{{\log _{{{\sin }^2}\frac{x}{2}}}\cos \frac{x}{2}}$ has the value equal to

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