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: 
**Hint.** Use the fact that $\ln u^v = v \ln u$ and the Taylor expansion for $\cos x$, $\sin x$; then use again the Taylor expansion of $\ln (1+x)$.
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?