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**Opalg** I don't know why you say that the function is discontinuous at 0. In fact, it is continuous at 0, because $\displaystyle \displaystyle\lim_{x\to0}\frac{\sin x}{1-e^{-x}} = 1$ (as you can check by using l'Hôpital's rule, for example).

To find the first few terms of the Taylor series, first write the first few terms for the series of the numerator and denominator:

$\displaystyle \displaystyle \frac{\sin x}{1-e^{-x}} = \frac{x - x^3/6 + \ldots}{x - x^2/2 + x^3/6 - \ldots} = \frac{1 - x^2/6 + \ldots}{1 - (x/2 - x^2/6 + \ldots)}.$

For the denominator, use the series $\displaystyle \frac1{1-z} = 1+z+z^2+\ldots$, with $\displaystyle z = x/2-x^2/6+\ldots$. Then, ignoring all powers greater than $\displaystyle x^2$, we get

$\displaystyle \displaystyle \frac{\sin x}{1-e^{-x}} = (1 - x^2/6 + \ldots)(1 + (x/2-x^2/6 + \ldots) + (x/2 - \ldots)^2).$

Now you just have to pick out the coefficients of the first few powers of x on the right side of that equation.