calculus - Prove $lim_{xto 0}frac{ln(cos x)}{lnleft(1-frac{x^2}{2}right)}=1$ 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.