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Prove that:$$\lim_{x\to 0}\frac{\ln(\cos x)}{\ln\left(1-\frac{x^2}{2}\right)}=1$$

without L'Hopital's rule.

I don't know if this is possible.

WolframAlpha agrees with this limit.

There's a similar limit without L'Hopital's rule $\lim_{x\to 0^+}\frac{\ln(\sin x)}{\ln (x)}=1$, which is easier to prove and some proofs can be seen in this question.

In $\lim_{x\to 0^+}\frac{\ln(\sin x)}{\ln (x)}=1$ you can see $x\to 0^+$, which is important, but in this case $\frac{\ln(\cos x)}{\ln\left(1-\frac{x^2}{2}\right)}$ is an even function and we can use $x\to 0$.

I've tried some of the methods given in the linked question and none of them worked.

I first saw this problem in this answer, where I also discussed the methods I've tried to solve this.

We could use $\lim_{x\to 0}\frac{\cos x}{1-\frac{x^2}{2}}=1$.


Edit: I've noticed that also

$$\lim_{x\to 0}\frac{\cos (\cos x)}{\cos\left(1-\frac{x^2}{2}\right)}=1$$

See WolframAlpha here.

And also $$\lim_{x\to 0}\frac{\sin (\cos x)}{\sin\left(1-\frac{x^2}{2}\right)}=1$$

See WolframAlpha here.

And also

$$\lim_{x\to 0}\frac{\tan (\cos x)}{\tan\left(1-\frac{x^2}{2}\right)}=1$$

See WolframAlpha here..

And also $$\lim_{x\to 0}\frac{\arcsin(\cos x)}{\arcsin\left(1-\frac{x^2}{2}\right)}=1$$

See WolframAlpha here.

Etc.

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  • $\begingroup$ All your limits after the fold (horizontal line) are trivial because putting $x=0$ does not create any problem. Same is the case with $(\cos x) /(1-(x^{2}/2))$. What is needed here are the limits $$\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2},\lim_{x\to 0}\frac{\log(1+x)}{x}=1$$ $\endgroup$
    – Paramanand Singh
    Sep 24, 2017 at 0:10
  • $\begingroup$ @ParamanandSingh Thanks. This gives an answer similar to Michael Rozenberg's. You could've posted this as an answer. I won't put the answer here to let you do it yourself. $\endgroup$
    – user263326
    Sep 24, 2017 at 5:44

4 Answers 4

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Using $\ln(1-y)=-\sum_{n=1}^\infty y^n/n$ we get immediately $\ln(1-x^2/2)=-x^2/2+O(x^4)$, and also, after observing that $\cos x=1-x^2/2+O(x^4)$, $\ln\cos x=-x^2/2+O(x^4)$. Therefore $$\frac{\ln \cos x}{\ln(1-x^2/2)}=1+O(x^2).$$

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  • $\begingroup$ You used $1-\cos x=x^2/2+O(x^4)$, $\ln(\cos x)=\ln(1-(1-\cos x))$, etc. You could use series division algorithm (similar to long division of polynomials) to find the series of $\frac{1}{\ln(1-x^2/2)}$ ($1=1+0x+0x^2+\cdots$) and then use Cauchy product to find the product of the series of $\ln(\cos x)$ and $\frac{1}{\ln(1-x^2/2)}$. I understand it. $\endgroup$
    – user263326
    Sep 23, 2017 at 19:42
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$$\frac{\ln{\cos{x}}}{\ln\left(1-\frac{x^2}{2}\right)}=\frac{\ln(1+\cos{x}-1)}{\cos{x}-1}\cdot\frac{-\frac{x^2}{2}}{\ln\left(1-\frac{x^2}{2}\right)}\cdot\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}\rightarrow1.$$

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  • $\begingroup$ Much better than my solution, +1 and I will delete my answer. $\endgroup$ Sep 23, 2017 at 18:55
  • $\begingroup$ Simple answer without any sophisticated tools. +1 $\endgroup$
    – Paramanand Singh
    Sep 23, 2017 at 23:45
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Hint : You can use the expansions to get the result

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On OP's request I am converting my comment into an answer. The limit in question is easily solved if one rewrites the given expression as $$\frac{\log(1+\cos x-1)} {\cos x - 1}\cdot\frac{\cos x - 1}{x^{2}}\cdot\frac{x^{2}}{(-x^{2}/2)}\cdot\frac{(-x^{2}/2)}{\log(1-x^{2}/2)}$$ And then the limit is easily seen to be $1\cdot(-1/2)\cdot(-2)\cdot 1=1$.

Other limits in your question (after the fold) don't have any problem as we can just plug $x=0$ to evaluate them (the functions concerned are continuous at $0$).

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  • $\begingroup$ A note: you could've also written $\frac{1-\cos x}{\frac{1}{2}x^{2}}$, which tends to $1$ as $x\to 0$, instead of $\frac{\cos x-1}{x^{2}}\cdot \frac{x^{2}}{(-x^{2}/2)}$. $\endgroup$
    – user263326
    Sep 24, 2017 at 6:05

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