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

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 1

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 agrees.

Answer 2

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 $\ln (1+x)$.

Answer 3

Like this:

$ L=\lim_{x \to 0} \frac{\ln \frac{\cos x}{1+\sin x}}{x}=\lim_{x \to 0}\frac{\frac{\cos x}{1+\sin x}-1}{x}=\lim_{x \to 0} \frac{\cos x-1-\sin x}{x(1+\sin x)}=\lim_{x \to 0}(-\sin x-\cos x)=-1 $

Answer 4

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?

Answer 5

For $1+\sin x\ne0,$ $$\dfrac{\cos x}{1+\sin x}=\dfrac{1-\tan\dfrac x2}{1+\tan\dfrac x2}=1+\dfrac{-2\tan\dfrac x2}{1+\tan\dfrac x2}$$

$$\lim_{x\to0}\left(\dfrac{\cos x}{1+\sin x}\right)^{1/x}$$

$$=\left(\lim_{x\to0}\left(1+\dfrac{-2\tan\dfrac x2}{1+\tan\dfrac x2}\right)^{-\dfrac{1+\tan\dfrac x2}{2\tan\dfrac x2}}\right)^{-\lim_{x\to0}\dfrac{\tan\dfrac x2}{\dfrac x2\left(1+\tan\dfrac x2\right)}}$$

Set $\dfrac{1+\tan\dfrac x2}{-2\tan\dfrac x2}=y$ to find the inner limit converges to $e$

Now find the limit of exponent