Finding a limit-value of a function

Questions
,
1

Hello, Colleagues

I have tried to find the limit value of a function.
f(x1,x2) = (x1^n + x2^n)^(1/n), as n->infinity.
As you know, if x1<x2, the limit value becomes x2.
Mathematica always gives x2 with a condition, and this condition is ’ if Log[x1]>0&&Log[x1]<Log[x2]&&Log[x2]>0’ which is different from x1<x2.
My question is whether it is possible to get x2 with the condition x1<x2.

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2

Not a full result, but this is what I was able to get:

Limit[(x1^n + x2^n)^(1/n), n -> \[Infinity], 
 Assumptions -> 0 < x1 < x2]
% /. ConditionalExpression[value_, cond_] :> 
  ConditionalExpression[value, 
   Reduce[cond && x1 > 0 && x2 > 0, Reals]]
3

There is an Assumption option in Limit or DiscreteLimit function:

In[43]:= DiscreteLimit[(x1^n+x2^n)^(1/n),n->\[Infinity],Assumptions->0<x1<x2<1]
Out[43]= x2
In[44]:= DiscreteLimit[(x1^n+x2^n)^(1/n),n->\[Infinity],Assumptions->1<x1<x2]
Out[44]= x2
4

Assuming Reals, then including Assumptions → x1 > 0 && x2 > 0 in this Limit would seem to be necessary in order for f to have defined values for any n.

Even with this amendment though, the result is not as expected (i.e., Max[x1, x2], or something equivalent, because by symmetry, if the limit is x2 when x1 < x2, then the limit is x1 when x2 < x1).

Since this isn’t happening, I think you’ve found a gap in Limit’s coverage.

5

Thanks for your idea.
May I ask a question further? Is there any way to get ‘x2’ without any condition utilizing ConditionalExpression?

6

Thanks for your idea.
I applied your method to another problem. But, this case provided the solution with conditions.

f[x1_, x2_, x3_] := {(x1 - x2)^n + (x2 - x3)^n + (x1 - x3)^n}^(1/n);
DiscreteLimit[f[x1, x2, x3], n -> \[Infinity], 
 Assumptions -> 0 < x1 < x2 < x3 < 1]
DiscreteLimit[f[x1, x2, x3], n -> \[Infinity], 
 Assumptions -> 1 < x1 < x2 < x3]

Q2.nb (8.77 KB)

7

Thanks for your suggestion.
Although using ‘Assuming Reals’, it cannot give the expected result, that is, Max[x1,x2].
I think there might be problems that Mathematica is unable to solve, which is represented as a gap as you said.