Q:

Demidovich 578, how does it prove?

Prove: $\lim\limits_{x \to 0} \frac{\ln \cosh x}{\ln \cos x}=-1$

A:

Hint:
$$\ln(b^2)=2\ln b$$
$$\cosh^2x=1+\sinh^2x$$
$$\cos^2x=1-\sin^2x$$
$$\lim_{h\to0}\dfrac{\ln(1+h)}h=1$$
$$\dfrac{\sinh x}x=\dfrac{e^{2x}-1}{2x}\cdot\dfrac1{e^x}=?$$

