= Solution
First expand the <exponential function> and divide by $x$:
$$
\frac{1-e^{-x}}x=1-\frac x2+\frac{x^2}6-\frac{x^3}{24}+O(x^4).
$$
Seek its reciprocal as $1+Ax+Bx^2+Cx^3+O(x^4)$. Equating coefficients in the product with this expression gives
$$
A-\frac12=0,\qquad B-\frac A2+\frac16=0,\qquad C-\frac B2+\frac A6-\frac1{24}=0.
$$
Hence $A=1/2$, $B=1/12$, $C=0$. The first three nonzero terms of the <Taylor expansion> are
$$
\boxed{\frac{x}{1-e^{-x}}=1+\frac x2+\frac{x^2}{12}+O(x^4).}
$$
Although the quotient is undefined at zero as written, its limit is one, and this <Taylor expansion> belongs to its continuous, <real analytic> extension.
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