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*}

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 $.
