Can I prove that $\frac{e^x - e^{\sin x}}{x - \sin x} \to 1$ as $x \to 0$ without using DLH? Can I prove that $$\frac{e^x - e^{\sin x}}{x - \sin x} \to 1$$ as $x \to 0$ without using DLH? 
UPDATE:
If you apply MVT at $f(u) = e^u$ at $[x,\sin x] \forall x>0$ and at $[x,\sin x] \forall x<0$ and use the squeeze theorem the limit is equal to 1.
 A: For small $x\geq 0$, we have $\sin(x)\leq x$. Let $I$ be the closed interval $I=[\sin(x),x]$ for fixed $x$. Define $f(x)=\exp(x)$. Then we know that $f$ is continous. By the mean value theorem, we have
\begin{align}
f'(\xi)  = \frac{f(x)-f(\sin(x))}{x-\sin(x)}
\end{align}
for $\xi \in I$
For $x\rightarrow 0$ it follows, that $\xi=0$ and thus $f'(\xi)=1$.
A: Here is the answer using Mean Value Theorem: $$e^x - e^{\sin x} = (x - \sin x)e^c \mbox{ for some } c \mbox{ between }x \mbox{ and }\sin x.$$  Now let $x \to 0$ and argue $e^c = 1$ as $x \to 0.$
A: The inequality of the geometric, logarithmic, and arithmetic means asserts that if $a\ne b$ and both are positive then
$$ \sqrt{ab} < \frac{a-b}{\log a-\log b} < \frac{a+b}2 $$
Take $a=e^x$ and $b=e^{\sin x}$.
A: This turns out to be too simple. Just note that if $y=\sin x, z=x-\sin x$, then both $y,z$ tend to $0$ with $x$. The given expression can be written as $$f(x)=\frac{e^{x}-e^{\sin x}}{x-\sin x}= e^{y}\cdot\frac{e^{z}-1}{z}$$ Now $e^{y}\to 1$ as $y\to 0$ and $(e^{z}-1)/z\to 1$ as $z\to 0$. Hence $f(x) \to 1$ as $x\to 0$.
