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 assumption is ignored.