Prove $\lim_{x\to 0}\frac{\ln(\cos x)}{\ln\left(1-\frac{x^2}{2}\right)}=1$ without L'Hopital 
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.
 A: 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
$$\frac{\ln \cos x}{\ln(1-x^2/2)}=1+O(x^2).$$
A: $$\frac{\ln{\cos{x}}}{\ln\left(1-\frac{x^2}{2}\right)}=\frac{\ln(1+\cos{x}-1)}{\cos{x}-1}\cdot\frac{-\frac{x^2}{2}}{\ln\left(1-\frac{x^2}{2}\right)}\cdot\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}\rightarrow1.$$
A: Hint : You can use the expansions to get the result
A: 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 $0$).
