For example, when I evaluate

Limit[(1 + x^n + (x^2/2)^n)^(1/n), n -> Infinity]

Mathematica does not output any result. When I evaluate

Plot[Limit[(1 + x^n + (x^2/2)^n)^(1/n), n -> Infinity], {x, -10, 10}]

a piecewise graph is output. But I want to get the function, not only the graph. How can I do it?

  • $\begingroup$ (x^/2)? Sure? $\endgroup$ Feb 5, 2013 at 3:43
  • $\begingroup$ The term (x^/2)^n)^(1/n) has a syntax error. Please edit your post to fix it. $\endgroup$
    – m_goldberg
    Feb 5, 2013 at 5:30

1 Answer 1


The short answer is you cannot (quite) do this. Limit has trouble with exp-log-polynomial growth at infinity in the situation where parameters are present.

You can give Limit assumptions about the parameter and that sometimes helps.

Limit[(1 + x^n + (x^2/2)^n)^(1/n), n -> Infinity, 
 Assumptions -> -1 < x < 1]

(* Out[1]= 1 *)

You can substitute a known but symbolic value, such as pi, for x. This is useful in cases where you believe (i) the limit will not involve that particular value and (ii) you believe the limit is some function of the parameter. For example, to get the quadratic growth part of that parametrized limit, you could do this.

Limit[(1 + x^n + (x^2/2)^n)^(1/n) /. x -> -Pi, n -> Infinity]

(* Out[6]= \[Pi]^2/2 *)

This does not prove anything but it gives a result that might all the same be useful: where the behavior is quadratic in x, it is x^2/2.

A graph suggests it is linear in x for 1<x<1. The value is simple x in this range. Here I use a symbolic-numeric value to indicate this.

Limit[(1 + x^n + (x^2/2)^n)^(1/n) /. x -> Pi/E, n -> Infinity]

(* Out[16]= \[Pi]/E *)

How to get Limit to do this better? I do not know.


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