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.

    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   *)

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   *)