How to tell Mathematica that a limit converges to a product of two variables

I have looked through the manual and Im simply just stuck trying to make Mathematica solve this:

Limit[
-m ((Abs[r + 1/2 λ k] – Abs[r – 1/2 λ k])/(Abs[r + 1/2 λ k] Abs[r – 1/2 λ k])),
λ -> 0]

I have renamed ℓ to λ, however I simply cannot figure out how to tell that the limit produces m l = μ.

Could someone please show me how this is done?

=================

=================

2 Answers
2

=================

Replace m by μ/λ to impose the condition m λ = μ.
I assume r = {x, y, z} and k = {0, 0, 1}; then you want Norm for vectors, since Abs is for complex/real numbers.

Block[{r = {x, y, z}, k = {0, 0, 1}},
Limit[-m ((Norm[r + 1/2 λ k] – Norm[r – 1/2 λ k]) /
(Norm[r + 1/2 λ k] Norm[r – 1/2 λ k])) /. m -> μ/λ,
λ -> 0] // Simplify[#, r ∈ Reals] &
]
(* -((z μ)/(x^2 + y^2 + z^2)^(3/2)) *)

If you want to put it in terms of r, it’s easiest just to retype it. Or if you must show Mathematica who’s the boss, try

Simplify[-((z μ)/(x^2 + y^2 + z^2)^(3/2)),
Norm[r]^2 == x^2 + y^2 + z^2 && Norm[r] >= 0]
(* -((z μ)/Norm[r]^3) *)

Note that Norm[r] >= 0 seems extraneous.

r = {x, y};
a = {a1, a2};
expr = 1/Sqrt[(r – a).(r – a)] – 1/Sqrt[(r + a).(r + a)];
(Series[expr, {a1, 0, 1}, {a2, 0, 1}] // Normal) /. x^2 -> R^2 – y^2 //
PowerExpand // Factor

(* (2 (a1 x + a2 y))/R^3 *)

Have fun!