Baffled with $\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$ 
Calculate $$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$$

Personal work:
$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^{\sin x}\cdot\cos x-\sin x}\cdot e^{\sin x}\over \sin x})=\cdots$$
This gets to nowhere. Also, I substituted $t=e^{\sin x}$ but I could not replace the $e^x$.

Use the third-order Maclaurin formulae
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3),
$$
$$
\sin x = x-\frac{x^3}{6}+o(x^3),
$$
$$
e^{\sin x} = e^{x-\frac{x^3}{6}+o(x^3)} = 
1+\left(x-\frac{x^3}{6}+o(x^3)\right)+\frac{1}{2} \left(x-\frac{x^3}{6}+o(x^3)\right)^2+
\frac16 \left(x-\frac{x^3}{6}+o(x^3)\right)^3 + o\left(\left( x-\frac{x^3}{6}+o(x^3) \right)^3 \right)= 
$$
$$
1+x-\frac{x^3}{6} + \frac{x^2}{2}+\frac{x^3}{6} + o(x^3);
$$
when expanding, we took into account only the summands up to the 3rd order, everything else gone into $o(x^3)$. Hence
$$
\frac{e^{\sin x}-e^x}{x-\sin x} = \frac{\frac{x^3}{6}+o(x^3)}{\frac{x^3}{6}+o(x^3)} = \frac{\frac16+o(1)}{\frac16+o(1)} \rightarrow 1.
$$
