Supposing that A is exactly a representable number then the smallest value greater than A that MATLAB can represent is A+eps(A). Keep in mind however that A+eps(A)/2 is probably going to round to A+eps(A)
Any value A+delta will round to something, either to A or to at least A+eps(A), and the function can be calculated on the rounded value.
Notice that I said above "Assuming that A is exactly a representable number". Most decimal numbers are not exactly representable. For example, 0.1 is not exactly representable.
Suppose that you had a function that was described mathematically as having a boundary at 1/10 and the right hand side involves log(A-1/10). Mathematically any value A strictly greater than 1/10 will have a positive subtraction and the log would be of a positive value. But with 1/10 exactly not being exactly representable you have to worry about whether there is any value that is mathematically less than 1/10 whose representation rounds upward and so mathematically A-1/10 would be negative but the representation of A rounds up so the comparison of representation_of_A to 1/10 might succeed. Would the representation minus 1/10 potentially give negative even though the comparison succeeded? The answer depends on exactly how the subtraction is coded. For example (10*A - 1)/10 is algebraically A-1/10 but due to rounding the calculated result could be different.
