Evaluating $\lim _{x\to 0}\left(\ln \left[\frac{1-x^2}{\ln \left(\cos \left(x\right)\right)}\right]\right)$ How do I evaluate the following limit?
\begin{align*}
\lim _{x\to 0}\left(\ln \left[\frac{1-x^2}{\ln \left(\cos \left(x\right)\right)}\right]\right)
\end{align*}
I tried it this way, but I'm not sure if it is correct.
\begin{align*}
\lim _{x\to 0}\left(\ln \left[\frac{1-x^2}{\ln \left(\cos \left(x\right)\right)}\right]\right) &=\lim _{x\to 0} \left( \ln[1-x^2]-\ln[\ln(\cos x)]\right)\\
&=\lim _{x\to 0}\ln(1-x^2)-\cos x\\
&=\ln(1)-1\\
&=-1
\end{align*}
 A: $$\ln(-|a|)=\ln |a|+i\pi$$
When $x$ is small but non-zero, we have $$0<\cos x < 1$$
$$\ln (\cos x) < 0$$
\begin{align}
\lim_{x \to 0} \ln (\ln (\cos(x)) &= \lim_{x \to 0}\ln (-\ln(\cos(x))  + i\pi
\\&= -\infty  + i \pi
\end{align}
A: I think I see the problem, OP is working under the false assumption that $\ln(x) \approx x$, however it's actually $\ln(1+x) \approx x$(*). Hence,
$$ \ln (\ln( \cos x) )= \ln \ln \big( 1+ (cos x -1) \big) \approx \ln ((\cos x -1) + \frac{ (1-cos x)^2}{2} + O( (1-cos x)^3 )$$
(*): ref
A: There is mistake in your solution as others have pointed out already.
Alternatively, You may proceed like this also:
Let $\displaystyle L=\ln\left(\frac{1-x^{2}}{\ln(\cos x)}\right)$
$\displaystyle \Longrightarrow e^{L} =\frac{1-x^{2}}{\ln(\cos x)} \Longrightarrow e^{-L} =\frac{\ln(\cos x)}{1-x^{2}} =\frac{\ln\left( 1-\sin^{2} x\right)}{\sin^{2} x} .\frac{\sin^{2} x}{2\left( 1-x^{2}\right)} =\frac{-1-o\left(\sin^{2} x\right)}{2\left( 1-x^{2}\right)} .\sin^{2} x$
It follows that: As $\displaystyle x\rightarrow 0$, we have $\displaystyle e^{-L}\rightarrow 0$, which is possible only when $\displaystyle L\rightarrow \infty $.
