Meaningless output of Limit

Question

The output of the command

Limit[D[(Exp[x]-1-x)/x^2,{x, n}],x ->0,Direction-> "FromAbove",Assumptions -> n > 2 &&
 n \[Element] Integers]

((DirectedInfinity[(-1)^n] + DirectedInfinity[(-1)^(1 + n)]) n!)/Gamma[2 + n]

is meaningless in view of

((DirectedInfinity[(-1)^n] + DirectedInfinity[(-1)^(1 + n)]) n!)/  Gamma[2 + n] /. n -> 3

Indeterminate

and

Limit[((DirectedInfinity[(-1)^n] + 
  DirectedInfinity[(-1)^(1 + n)]) n!)/Gamma[2 + n], n -> 3]

which returns the input.

It is clearly understood that the result of

FullSimplify[FindSequenceFunction[Table[Limit[D[(Exp[x] - 1 - x)/x^2, {x, n}], x -> 0, 
Direction -> "FromAbove"], {n, 3, 6}], n] /. n -> n - 2]

1/(2 + 3 n + n^2)

is only a guess, not a workaround. We still have no certainty about the values of the limit under consideration for astronomic values of n (BTW, FindSequenceFunction[{-3, 5, -7, 9, -11}, n] returns the input.).

Is there a relaible workaround for this limit?

Answer 1

The limit of the $n$th derivative of a function at zero is equal to $n!$ times the $n$th coefficient in the functions Maclaurin series:

limit[n_] := n! SeriesCoefficient[(Exp[x] - 1 - x)/x^2, {x, 0, n}]
Simplify[limit[n], Assumptions -> n >= 2 && n \[Element] Integers]

(* n!/(2 + n)! *)

Why Mathematica doesn't simplify this further to 1/((n+1)(n+2)) I can't say, but it's definitely a more useful form than what Limit yields.

Answer 2

Using this command

Assuming[n > 2 && Element[n, Integers], 
 Limit[D[(Exp[x] - 1 - x)/x^2, {x, n}] // Simplify, x -> 0] //FunctionExpand]

I am getting

$$\frac{1}{(n+1) (n+2)}.$$