3 added 26 characters in body edited Apr 12 '16 at 6:28 J. M. will be back soon♦ 101k1010 gold badges317317 silver badges477477 bronze badges How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] (* {2} *)  Or a more direct solution (with thanks to xzczd): Select[x /. Solve[Denominator[r[x]] == 0, x],  Head@Residue[r[x] FreeQ[Residue[r[x], {x, #}] =!=, Residue_Residue] &]  How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] (* {2} *)  Or a more direct solution: Select[x /. Solve[Denominator[r[x]] == 0, x], Head@Residue[r[x], {x, #}] =!= Residue &]  How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] {2}  Or a more direct solution (with thanks to xzczd): Select[x /. Solve[Denominator[r[x]] == 0, x],  FreeQ[Residue[r[x], {x, #}], _Residue] &]  2 added 134 characters in body edited Apr 12 '16 at 2:49 xzczd 29.6k66 gold badges8383 silver badges273273 bronze badges How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] (* {2} *)  Or a more direct solution: Select[x /. Solve[Denominator[r[x]] == 0, x], Head@Residue[r[x], {x, #}] =!= Residue &]  How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] {2}  How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] (* {2} *)  Or a more direct solution: Select[x /. Solve[Denominator[r[x]] == 0, x], Head@Residue[r[x], {x, #}] =!= Residue &]  1 answered Mar 13 '16 at 2:15 J. M. will be back soon♦ 101k1010 gold badges317317 silver badges477477 bronze badges How about r[x_] := 1/((x - 2) Sqrt[x - 1]) Select[x /. Solve[Denominator[r[x]] == 0, x], SeriesCoefficient[r[x], {x, #, -1}] != 0 &] {2}  Post Made Community Wiki by J. M. will be back soon♦ occurred Mar 13 '16 at 2:15