2
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ If `f`` is not continuous, the limit could be anything. $\endgroup$ Commented Feb 4, 2021 at 17:07
  • $\begingroup$ Yes, that's why I'm wondering if there's a way to get it to assume f is continuous. $\endgroup$ Commented Feb 4, 2021 at 17:11
  • $\begingroup$ 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. $\endgroup$ Commented Feb 4, 2021 at 17:21
  • $\begingroup$ 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? $\endgroup$ Commented Feb 4, 2021 at 17:24
  • 1
    $\begingroup$ 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"]} $\endgroup$
    – Bob Hanlon
    Commented Feb 4, 2021 at 18:36

1 Answer 1

5
$\begingroup$

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]*)
$\endgroup$
1
  • $\begingroup$ Awesome! Thank you so much. $\endgroup$ Commented Feb 4, 2021 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.