$\lim_{x \to 0} \frac{e^{\sin2x}-e^{\sin x}}{x}$ 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?

Hint: Let $f(x)=e^{\sin(2x)}-e^{\sin(x)}$, for all $x\in \mathbb R$.
Note that $f(0)=0$ and think about the definition of $f'(0)$.
