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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?

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    $\begingroup$ Yes it seems right! Well done. $\endgroup$
    – user
    Aug 31, 2018 at 20:17

3 Answers 3

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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 $4$.

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  • $\begingroup$ Is this a standard result or something you just know? $\endgroup$
    – prog_SAHIL
    Aug 31, 2018 at 20:07
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    $\begingroup$ No, it is i not a standard result. It follows from Taylor expansions . $\endgroup$
    – Robert Z
    Aug 31, 2018 at 20:11
  • $\begingroup$ Nice generalization! $\endgroup$
    – user
    Aug 31, 2018 at 20:17
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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

$$\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{(x/2)}}{\cos{(x/2)}}} =\frac{\log{(1-\sin^2{x})}}{ {\log{(1-\sin^2{(x/2)})}}} \frac{\log {\sin{(x/2)}}}{\log\sin{x}}=$$

$$=\frac{\log{(1-\sin^2{x})}}{-\sin^2 x} \frac{-\sin^2 (x/2)}{ {\log{(1-\sin^2{(x/2)})}}} \frac{\sin^2 x}{x^2} \frac{4(x/2)^2}{\sin^2 (x/2)} \frac{\log {\sin{(x/2)}}}{\log 2+\log\sin{(x/2)}+\log \cos (x/2)}$$$$\to 1\cdot 1\cdot 1 \cdot 4 \cdot 1=4$$

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  • $\begingroup$ A very nice solution indeed. It is easy to solve. $\endgroup$
    – prog_SAHIL
    Sep 1, 2018 at 3:33
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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.

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  • $\begingroup$ Will you please try solving the second limit without L-Hopital? I liked your expansion method for the first one. +1 $\endgroup$
    – prog_SAHIL
    Sep 1, 2018 at 3:29
  • $\begingroup$ @prog_SAHIL This boils down to $\lim_{x\to0^+}\frac{\log\sin x}{\log x}=1$, which is a good example for l’Hôpital, in my opinion. $\endgroup$
    – egreg
    Sep 1, 2018 at 6:23

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