Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$ 
Calculate
  $$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$

My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{x}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\frac{-2\sin^2{\frac{x}{2}}}{-2\sin^2{\frac{x}{2}}}}\cdot\frac{\frac{-2\sin^2{\frac{x}{4}}}{-2\sin^2{\frac{x}{4}}}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{\frac{x}{2}}}+\ln{2}}$$
$$\frac{1}{4}\cdot 16\cdot 1$$
$$=4$$
Am I solving this correct?
 A: You are correct. This is another way to proceed: if $x\to 0^+$ then by using Taylor expansions we obatin
$$f(x):=\log_{\sin(x)}{(\cos (x))}=\frac{\log(\cos(x))}{\log(\sin(x))}=\frac{\log(1-\frac{x^2}{2}+o(x^2))}{\log (x+o(x^2))}\sim-\frac{x^2}{2\log(x)},$$
and it follows that for any $a>0$,
$$\lim_{x\to{0^+}}\frac{\log_{\sin(x)}{\cos(x)}}{\log_{\sin(ax)}\cos(ax)}=\lim_{x\to{0^+}}\frac{f(x)}{f(ax)}=\lim_{x\to{0^+}}\frac{-\frac{x^2}{2\log(x)}}{-\frac{(ax)^2}{2\log(ax)}}=\lim_{x\to{0^+}}\frac{\log(x)+\log(a)}{a^2\log(x)}=\frac{1}{a^2}.$$
In your case $a=1/2$ and the limit is $4$.
A: Yes it looks fine, as a similar alternative we can use
$$\log_{\sin{x}}{\cos{x}}=\frac12\log_{\sin{x}}{\cos^2{x}}=\frac12\log_{\sin{x}}{(1-\sin^2{x})}=\frac12\frac{\log{(1-\sin^2{x})}}{\log {\sin{x}}}$$
$$\log_{\sin{(x/2)}}{\cos{(x/2)}}=\frac12\frac{\log{(1-\sin^2{(x/2)})}}{\log {\sin{(x/2)}}}$$
then
$$\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{(x/2)}}{\cos{(x/2)}}}
=\frac{\log{(1-\sin^2{x})}}{ {\log{(1-\sin^2{(x/2)})}}}
\frac{\log {\sin{(x/2)}}}{\log\sin{x}}=$$
$$=\frac{\log{(1-\sin^2{x})}}{-\sin^2 x}
\frac{-\sin^2 (x/2)}{ {\log{(1-\sin^2{(x/2)})}}}
\frac{\sin^2 x}{x^2}
\frac{4(x/2)^2}{\sin^2 (x/2)}
\frac{\log {\sin{(x/2)}}}{\log 2+\log\sin{(x/2)}+\log \cos (x/2)}$$$$\to 1\cdot 1\cdot 1 \cdot 4 \cdot 1=4$$
A: You can rewrite the function as
$$
\frac{\log\cos(x)}{\log\cos(x/2)}\frac{\log\sin(x/2)}{\log\sin(x)}
$$
The first fraction is easy to deal with:
$$
\log\cos x=\log(1-x^2/2+o(x^2))=-\frac{x^2}{2}+o(x^2)
$$
and similarly
$$
\log\cos\frac{x}{2}=\log(1-x^2/8+o(x^2))=-\frac{x^2}{8}+o(x^2)
$$
Therefore
$$
\lim_{x\to0}\frac{\log\cos(x)}{\log\cos(x/2)}=4
$$
For the second fraction, it's simpler to consider $x=2t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{\log\sin(t)}{\log\sin(2t)}=
\lim_{t\to0^+}\frac{\dfrac{\cos(t)}{\sin(t)}}{\dfrac{2\cos(2t)}{\sin(2t)}}=
\lim_{t\to0^+}\frac{\cos(t)}{\cos(2t)}\frac{\sin(2t)}{2\sin(t)}=1
$$
with a simple application of l’Hôpital.
