Evaluate $\lim\limits_{x\to0}\frac{\log_{\sin^2x}\cos x}{\log_{\sin^2{\frac x2}}\cos\frac x2}$ 
$$\lim\limits_{x\to0}\frac{\log_{\sin^2x}\cos x}{\log_{\sin^2{\frac x2}}\cos\frac x2}$$

I tried writing the denominator as $\frac{\log_{\sin^2{x}}\cos\frac x2}{\log_{\sin^2x}\sin^2\frac x2}$ but couldn't see further. Moreover, the first thought that came to my mind was applying the L'Hopital's rule, but we are not getting a $\frac00$ form, rather, we are getting a $\frac{\text{undefined}}{\text{undefined}}$ form. (Logarithms to the base $0$ are undefined). 

So what can I do forward?
 A: Hint. You may use the fact that, by the Taylor expansion, as $x\to 0$,
$$
\cos (ax)=1-a^2\frac{x^2}{2!}+O(x^4) \tag1
$$
$$
\sin (ax)=ax-a^3\frac{x^3}{3!}+O(x^4) \tag2
$$
and, as $u\to 0$, 
$$
\log(1-u)=-u+O(u^2) \tag3
$$ 
giving
$$
\log(\cos(ax))=-\frac{a^2 x^2}{2}+O(x^4) \tag4
$$$$
\log(\sin(ax))=\log(ax)-\frac{a^2 x^2}{6}+O(x^4) \tag5
$$ to get, as $x \to 0$,

$$
\frac{\log_{\sin^2(ax)}\cos(a x)}{\log_{\sin^2{(bx)}}\cos (bx)}=\frac{\log(\cos(a x))}{\log(\sin^2 (ax))}\times\frac{\log(\sin^2 (bx))}{\log(\cos (bx))}=\frac{a^2}{b^2}+O(1/\log(x)) \tag6
$$ 

then obtain the desired limit with $a=1$ and $b=1/2$.
A: Thanks to Jan Eerland, who suggested to write the original expression as
\begin{equation*}
\frac{\log _{\sin ^{2}x}\left( \cos x\right) }{\log _{\sin ^{2}\left( \frac{x%
}{2}\right) }\left( \cos \left( \frac{x}{2}\right) \right) }=\frac{\frac{\ln
\left( \cos x\right) }{\ln \left( \sin ^{2}x\right) }}{\frac{\ln \left( \cos
(x/2)\right) }{\ln \left( \sin ^{2}(x/2)\right) }}=\frac{\ln (\cos x)\ln
(\sin ^{2}(\frac{x}{2}))}{\ln (\cos \frac{x}{2})\ln (\sin ^{2}(x))}.
\end{equation*}
Now, it remains just to transform the last expression as a product of other
expressions whose limits are known or maybe computed easily. For example, $
\frac{\ln (\cos x)\ln (\sin ^{2}(\frac{x}{2}))}{\ln (\cos \frac{x}{2})\ln
(\sin ^{2}(x))}$
\begin{eqnarray*}
&=&\frac{\ln (1-\sin ^{2}x)\ln (\sin ^{2}(\frac{x}{2}))}{\ln (1-\sin ^{2}%
\frac{x}{2})\ln (\sin ^{2}(x))} \\
&=&\left( \frac{\ln (1-\sin ^{2}x)}{\sin ^{2}x}\right) \left( \frac{\sin
^{2}(\frac{x}{2})}{\ln (1-\sin ^{2}\frac{x}{2})}\right) \frac{\ln (\sin ^{2}(%
\frac{x}{2}))}{\ln (\sin ^{2}(x))}\left( \frac{\sin x}{x}\right) ^{2}\left( 
\frac{\left( \frac{x}{2}\right) }{\sin \left( \frac{x}{2}\right) }\right)
^{2}\left( \frac{x}{\left( \frac{x}{2}\right) }\right) ^{2} \\
&=&\left( \frac{\ln (1-\sin ^{2}x)}{\sin ^{2}x}\right) \left( \frac{\sin
^{2}(\frac{x}{2})}{\ln (1-\sin ^{2}\frac{x}{2})}\right) \left( \frac{\ln
(\sin (\frac{x}{2}))}{\ln (\sin (\frac{x}{2})\left( 2\cos \left( \frac{x}{2}%
\right) \right) )}\right) \left( \frac{\sin x}{x}\right) ^{2}\left( \frac{%
\left( \frac{x}{2}\right) }{\sin \left( \frac{x}{2}\right) }\right)
^{2}\times 4 \\
&=&\left( \frac{\ln (1-\sin ^{2}x)}{\sin ^{2}x}\right) \left( \frac{\sin
^{2}(\frac{x}{2})}{\ln (1-\sin ^{2}\frac{x}{2})}\right) \left( \frac{1}{1+%
\frac{\ln \left( 2\cos \left( \frac{x}{2}\right) \right) }{\ln (\sin (\frac{x%
}{2}))}}\right) \left( \frac{\sin x}{x}\right) ^{2}\left( \frac{\left( \frac{%
x}{2}\right) }{\sin \left( \frac{x}{2}\right) }\right) ^{2}\times 4.
\end{eqnarray*}
Therefore,
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\log _{\sin ^{2}x}\left( \cos x\right) }{\log
_{\sin ^{2}\left( \frac{x}{2}\right) }\left( \cos \left( \frac{x}{2}\right)
\right) }=\left( -1\right) \times \left( -1\right) \times \left( \frac{1}{1+0%
}\right) \times \left( 1\right) ^{2}\times \left( 1\right) ^{2}\times 4=4.
\end{equation*}
A: HINT:
$$\lim\limits_{x\to0}\frac{\log_{\sin^2x}\cos x}{\log_{\sin^2{\frac x2}}\cos\frac x2}=$$
$$\lim_{x\to 0} \frac{\frac{\ln\left(\cos(x)\right)}{\ln\left(\sin^2(x)\right)}}{\frac{\ln\left(\cos\left(\frac{x}{2}\right)\right)}{\ln\left(\sin^2\left(\frac{x}{2}\right)\right)}}=$$
$$\lim_{x\to 0} \frac{\ln\left(\cos(x)\right)\ln\left(\sin^2\left(\frac{x}{2}\right)\right)}{\ln\left(\cos\left(\frac{x}{2}\right)\right)\ln\left(\sin^2(x)\right)}$$
