$\lim_{x \to 0} \frac{e^{\sin2x}-e^{\sin x}}{x}$ without L'Hospital Anyone has an idea how to find $\displaystyle\lim_{x \to 0} \dfrac{e^{\sin2x}-e^{\sin x}}{x}$ without L'Hospital? I solved it with L'Hospital and the result is $1$ but the assignment is to find it without L'Hospital. Any idea?

$$\eqalign{
\lim_{x \to 0} \dfrac{e^{\sin2x}-e^{\sin x}}{x}
&=\lim_{x \to 0}\left(\dfrac{1}{x}+\dfrac{\sin2x}{x}+\dfrac{\sin^22x}{2x}+\cdots-\dfrac{1}{x}-\dfrac{\sin x}{x}-\dfrac{\sin^2 x}{2x}-\cdots\right)\\
&=\lim_{x \to 0}\left(\dfrac{\sin2x}{x}+\dfrac{\sin^22x}{2x}+\cdots-\dfrac{\sin x}{x}-\dfrac{\sin^2 x}{2x}-\cdots\right)\\
&=\left(2+0+\cdots-1-0-\cdots\right)\\
&=2-1=1}$$
