\displaystyle\lim_{x\to 0}\frac{e^{sin(x)}}{e^x}\frac{\frac{e^{sin(2x)-sin(x)}-1}{sin(2x)-sin(x)}}{\frac{e^x-1}{x}}(2\frac{sin(2x)}{2x}-\frac{sin(x)}{x})=\frac{1}{1} \cdot \frac{1}{1} \cdot (2 \cdot 1-1)=1