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

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?


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*}


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 $b=1/2$.



$$\lim\limits_{x\to0}\frac{\log_{\sin^2x}\cos x}{\log_{\sin^2{\frac x2}}\cos\frac x2}=$$ $$\lim_{x\to 0} \frac{\frac{\ln\left(\cos(x)\right)}{\ln\left(\sin^2(x)\right)}}{\frac{\ln\left(\cos\left(\frac{x}{2}\right)\right)}{\ln\left(\sin^2\left(\frac{x}{2}\right)\right)}}=$$ $$\lim_{x\to 0} \frac{\ln\left(\cos(x)\right)\ln\left(\sin^2\left(\frac{x}{2}\right)\right)}{\ln\left(\cos\left(\frac{x}{2}\right)\right)\ln\left(\sin^2(x)\right)}$$

  • $\begingroup$ Now L'Hopital? Or any other easier and less messy method exists? $\endgroup$ – Aditya Agarwal Oct 10 '15 at 11:01
  • $\begingroup$ You should reply to the comments. Bad behavior. @JanEerland $\endgroup$ – Aditya Agarwal Oct 10 '15 at 12:57
  • $\begingroup$ Stfu, I gave you a hint! $\endgroup$ – Jan Oct 10 '15 at 13:14

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