Question: How to evaluate this limit?

$$\lim\_{x \to 0} \ln\left(\frac{\cos x}{1+\sin x}\right)^{\frac{1}{x}}$$

I think, somehow it can be transformed to expression, similar to $\frac{\ln (1+x)}{x}$
which limit is equal to $1$, but how?
Can you help?$

Answer: 
Using L'Hopital's Rule, we have
\begin{align}
\lim\_{x\to0}\ln\left(\frac{\cos x}{1+\sin x}\right)^{1/x}&=\lim\_{x\to0}\frac{\ln(\cos x)-\ln(1+\sin x)}{x}\\
&=\lim\_{x\to0}\left(-\tan x-\frac{\cos x}{1+\sin x}\right)\\
&=-1
\end{align}

[WolframAlpha](https://www.wolframalpha.com/input/?i=limit+x+to+0+ln%28%28cosx%2F%281%2Bsinx%29%29%5E%281%2Fx%29%29) agrees.
We have
\begin{aligned}
\lim\limits\_{x\to 0}\ln\left(\frac{\cos x}{1+\sin x}\right)^{\frac{1}{x}}&=\lim\limits\_{x\to 0}\left(\frac{\ln(1+(\cos x-1))-\ln(1+\sin x)}{x}\right)\\
&=\lim\limits\_{x\to 0}\left(\frac{\ln(1+(\cos x-1))}{x(\cos x-1)}\cdot\frac{\cos x-1}{x}\cdot x-\frac{\ln(1+\sin x)}{x\sin x}\cdot \sin x\right)\end{aligned}

Can you continue?