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I'm trying to use this function in mathematica 11:

differentiableAtQ[
  f_, p_?VectorQ, vars_?VectorQ, dom_ : Reals
  ] := With[{n = Length[vars], dimP = Length[p]},
  If[n < 1 || n != dimP, Return[]];
  If[n > 1,
   With[{pd = D[f, #] & /@ vars},
    With[{pdValues = ((Evaluate[vars] \[Function] #) @@ p) & /@ pd},
     (* All partial derivatives exist *)
     AllTrue[pdValues, NumericQ] &&
      With[{$f = Evaluate[vars] \[Function] Evaluate[f]},
       (* All partial derivatives are continuous *)
       AllTrue[{pd, pdValues}\[Transpose],
         Apply[Limit[#1, vars -> p] === #2 &]
         ] || Switch[ (* Taking limit *)
         Limit[FullSimplify[
           (If[MemberQ[#, _Piecewise, \[Infinity]],
               # // PiecewiseExpand, #] &)[
            (* Edit for correction (n-1) *)
            ($f @@ vars + (n - 1) $f @@ p
               - Total[
                $f @@@ (ConstantArray[vars, n]
                   + DiagonalMatrix[p - vars])
                ])/Norm[vars - p]],
           And @@ Thread[vars != p]
            && vars \[Element] dom],
          vars -> p],
         0, True,
         Indeterminate, False,
         _DirectedInfinity, False,
         _, Indeterminate
         ]
       ]]],
   D[f, vars] /. vars[[1]] -> p[[1]] // NumericQ
   ]]

copied from this question How to write a custom function to judge whether a bivariate function is differentiable at a certain point?

Maybe in mathematica 12 it works fine, but in 11 if I test it with some examples taken from the same post, it say:

General : {x,y} is not a valid variable .

So, how can I adapt this function to make it usable in mathematica 11?

Improve this question
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    $\begingroup$ The problem here is that Limit only gained the functionality for multi-dimensional limits in version 11.2, so in version 11 you can only consider the limit along parametrized paths. I am not sure how exactly to replace this functionality, since multi-dimensional limits are not easy to compute, as existence of the limit along a path does not prove the multi-variate limit exists (check e.g. this question) $\endgroup$
    – Hausdorff
    Commented Aug 30, 2020 at 22:15
  • 1
    $\begingroup$ For 2D you could try a shrinking disk using a polar substitution f[x+x0, y+y0] /. {x -> r Cos[θ], y -> r Sin[θ]} and then take the limit as r->0. If the result is dependent on θ or not a constant function then the limit is indeterminate, otherwise it should work. The same should work for a shrinking sphere in 3D. ^ Note: this assumes the f is continuous on the disks everywhere except the point x0,y0 $\endgroup$
    – flinty
    Commented Aug 30, 2020 at 23:08

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