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

$e^{sin2x} \sim_0 sin2x + 1$, and also $e^{sinx} \sim_0 sinx + 1$. Thus:
$\dfrac{e^{sin2x} - e^{sinx}}{x} \sim_0 \dfrac{sin2x + 1 - sinx - 1}{x} \sim_0 \dfrac{2x - x}{x} = \dfrac{x}{x} = 1$
