# Can Limit take limits of unknown functions?

For example, when I input this

 f[0] = 0;
Limit[f[x], x -> 0]


, it does not output 0.

Is there a way to get it to output 0 in this case? For example, is there a way I can get Mathematica to assume f is continuous?

• If f is not continuous, the limit could be anything. Commented Feb 4, 2021 at 17:07
• Yes, that's why I'm wondering if there's a way to get it to assume f is continuous. Commented Feb 4, 2021 at 17:11
• It is sort of a chicken and egg question. To assume that the function is continuous in zero is the same as saying that both limits are zero. Commented Feb 4, 2021 at 17:21
• I agree, but ultimately I would like to do something more complicated, for example if f and g are both C1 at 0 and g'[0] != 0, to take Limit[f[x]/g[x],x->0]. Is there no way to give mathematica the assumption functions are continuous? Commented Feb 4, 2021 at 17:24
• Define f /: Limit[f[z_], z_ -> 0, ___] := 0 then all limits at 0 are 0. {Limit[f[x], x -> 0], Limit[f[x], x -> 0, Direction -> "FromAbove"], Limit[f[x], x -> 0, Direction -> "FromBelow"], Limit[f[x], x -> 0, Direction -> "Reals"], Limit[f[x], x -> 0, Direction -> "TwoSided"], Limit[f[x], x -> 0, Direction -> 1], Limit[f[x], x -> 0, Direction -> -1], Limit[f[x], x -> 0, Direction -> "Complexes"]} Commented Feb 4, 2021 at 18:36

Yes, it is possible.

ClearAll["Global*"];
Limit[f[x], x -> 0,Analytic->True]
(*f[0]*)


Unfortunately,this option does not work for multidimensional limits, but works for iterated limits, e.g.

Limit[g[x, y], {x -> 1, y -> 0}, Analytic -> True]
(* g[1,0]*)

• Awesome! Thank you so much. Commented Feb 4, 2021 at 20:07