Do integration first and then take the limit k-> inf.
Use indefinite integration. Integrate only finds a solution for m==0.
g[x_, m_, k_] = 1/((2*(x - m))^(2*k) + 1)
mint[x_, k_] = Integrate[g[x, 0, k], x]
(* x Hypergeometric2F1[1, 1/(2 k), 1 + 1/(2 k), -4^k x^(2 k)] *)
Use a trick. Tell Limit that x^(2 k) is always positive. (Valid for Integer k)
That is only neccessary for x== -1/2. For all other positive and negative x Limit[mint[x, k], k -> \[Infinity]] is x. Maybe this is a removable singularity at x== -1/2 ?
mint2[x_, k_] = mint[x, k] /. x^(2 k) -> Abs[x]^(2 k)
Limit[mint2[1/2, k] - mint2[-1/2, k], k -> \[Infinity]]
(* 1 *)
A graphic shows the same.
Manipulate[Plot[mint[x, k], {x, -.5, .5}], {k, 1, 1000}]
Rubi (https://rulebasedintegration.org/) does the integral with arbitrary m.
rint[x_, m_, k_] = Int[g[x, m, k], x]
(* (-m + x) Hypergeometric2F1[1, 1/(2 k),
1/2 (2 + 1/k), -4^k (-m + x)^(2 k)] *)
rint2[x_, m_, k_] =
rint[x, m, k] /. (-m + x)^(2 k) -> Abs[(-m + x)]^(2 k)
Limit[rint2[m + 1/2, m, k] - rint2[m - 1/2, m, k], k -> \[Infinity]]
(* 1 *)