3
$\begingroup$

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

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?

$\endgroup$
4
$\begingroup$

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*}

$\endgroup$
5
$\begingroup$

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 $b=1/2$.

$\endgroup$
1
$\begingroup$

HINT:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.