$\lim_{x \to 0} \frac{e^{\sin2x}-e^{\sin x}}{x}$ 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?
 A: I have more elementary way.
$$\lim_{x \rightarrow 0}\frac{e^{\sin 2x}-e^{\sin x}}{x}=\lim_{x \rightarrow 0}\frac{e^{2\sin x\cos x}-e^{\sin x}}{x}=\lim_{x \rightarrow 0}\frac{e^{\sin x}(e^{2\sin x\cos x - \sin x}-1)}{x}=$$
$$\lim_{x \rightarrow 0}\frac{e^{\sin x(2\cos x -1)}-1}{\sin x(2\cos x -1)}\cdot \frac{\sin x}{x}\cdot e^{\sin x}(2\cos x-1)=1\cdot1\cdot1\cdot(2-1)=1 $$
Using the limit: 
$$\lim_{x \rightarrow 0}\frac{e^x-1}{x}=1$$
A: Note that
$$\lim_{x \to 0} \frac{e^{\sin 2x} - e^{\sin x}}{x} = \lim_{x \to 0} \frac{e^{\sin 2x} - e^{\sin x}}{\sin x}$$
after using the fact that $\sin x / x \to 1$ as $x \to 0$. Now from the identity $\sin 2x = 2 \sin x \cos x$, and making the substitution $t = \sin x$, we want to compute
$$\lim_{t \to 0} \frac{e^{2t \sqrt{1 - t^2}} - e^t}{t}$$
Now  by a Taylor expansion, the numerator is
$$2t \sqrt{1 - t^2} - t + o(t^2)$$
Dividing by $t$ now means that we want to compute
$$\lim_{t \to 0} 2 \sqrt{1 - t^2} - 1 = 1$$
A: $e^{sin2x} \sim_0 sin2x + 1$, and also $e^{sinx} \sim_0 sinx + 1$. Thus:
$\dfrac{e^{sin2x} - e^{sinx}}{x} \sim_0 \dfrac{sin2x + 1 - sinx - 1}{x} \sim_0 \dfrac{2x - x}{x} = \dfrac{x}{x} = 1$
A: $$\eqalign{
\lim_{x \to 0} \dfrac{e^{\sin2x}-e^{\sin x}}{x}
&=\lim_{x \to 0}\left(\dfrac{1}{x}+\dfrac{\sin2x}{x}+\dfrac{\sin^22x}{2x}+\cdots-\dfrac{1}{x}-\dfrac{\sin x}{x}-\dfrac{\sin^2 x}{2x}-\cdots\right)\\
&=\lim_{x \to 0}\left(\dfrac{\sin2x}{x}+\dfrac{\sin^22x}{2x}+\cdots-\dfrac{\sin x}{x}-\dfrac{\sin^2 x}{2x}-\cdots\right)\\
&=\left(2+0+\cdots-1-0-\cdots\right)\\
&=2-1=1}$$
A: 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 $f'(0)$.
