Question

Consider the following two lines of code, which differ only by replacing 1 by y:

SeriesCoefficient[
  (Exp[x + 1 + a] - 1)/(x + 1 + a), {a, 0, 0}, Assumptions -> x + 1 == 0]

SeriesCoefficient[
  (Exp[x + y + a] - 1)/(x + y + a), {a, 0, 0}, Assumptions -> x + y == 0]

The first version yields 1, which is the correct answer (just the limit of the function as "a" goes to zero). The second version yields:

(-1 + Exp[x + y])/(x + y)

which will cause problems for me later because x + y == 0.

Assuming x and y are real does not change the result.

Background

I want to take the limit of a large number of continuous functions of several variables:

Limit[
  Limit[Limit[Limit[f[y1, y2, y3, y4], y1 -> x1], y2 -> x2], y3 -> x3],
  y4 -> x4]

where each function f is similar to the one used above (ie it naively seems to have singularities but is actually continuous). I obtain much faster results using SeriesCoefficient:

SeriesCoefficient[f[y1 + a, y2 + a, y3 + a, y4 + a], {a, 0, 0}]

I want to be able to evaluate this with assumptions. For example:

SeriesCoefficient[
  f[x1 + a, x2 + a, x3 + a, x4 + a], {a, 0, 0}, 
  Assumptions -> x1 + x2 + x3 + x4 == 0]

but the assumptions are ignored.