\displaystyle \lim_{x \to 0} {\frac{\ln \left( 1+e^x-e^ {2x}\right)}{\sin x-\textrm{sh} 2x}}=\lim_{x \to 0} {\frac{\ln \left( 1+e^x-e^ {2x}\right)}{e^x-e^ {2x}}}\cdot \left( \frac{e^x-1}{\sin x-\textrm{sh} 2x} + \frac{1-e^{2x}}{\sin x-\textrm{sh} 2x}\right)=