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 $e^x$.

You are exactly one application of L'Hospital away from getting the answer, because the next denominator will be (non-vanishing!) $\cos x$. Alternatively, we can see that three applications of L'Hospital's rule are necessary because $x - \sin x$ vanishes to third order at $0$.

Here's an approach using series with a detailed tracking of the error terms. We only need to keep terms up to the cubics, so we have 
$$\frac{1 + x + x^2 / 2 + x^3/6 + O(x^4) - \big(1 + \sin x + \sin^2 x / 2 + \sin^3 x / 6 + O(\sin^4 x)\big)}{x - \big(x - x^3/6 + O(x^5)\big)}$$
which is equal to
$$\frac{x + x^2/2 + x^3 / 6 + O(x^4) - (x - x^3/6 + O(x^5) + \frac 1 2(x^2 + O(x^4)) +  x^3/6 + O(x^4)}{x^3 / 6 + O(x^5)}$$
which in turn simplifies as
$$\frac{x^3 / 6 + O(x^4)}{x^3/6 + O(x^5)} \to 1.$$
