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I'm trying to determine the limit of the sequence of functions

$$f_n(x)=\left(\frac{1}{\pi}\arctan(n x) + 1/2\right)^n. $$

I define

f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n

And enter

Limit[f[x, n], n -> Infinity]

This gives the answer 0.

If I instead enter

Assuming[x > 0, Limit[f[x, n], n -> Infinity]]

I get the answer $$e^{-\frac{1}{x\pi}}. $$

While

Assuming[x < 0, Limit[f[x, n], n -> Infinity]]

gives the answer 0 again.

Why does the normal Limit give the answer assuming that x<0? Is this a bug, or have I missed something?

Thanks in advance.

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  • $\begingroup$ Please only use the tag bugs post-hoc, if it is confirmed by WRI or the community. $\endgroup$ Oct 16 '13 at 12:19
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    $\begingroup$ I think it's a bug. The result is not always 0 as you check by using 1 for instance in place of x. $\endgroup$ Oct 16 '13 at 12:51
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    $\begingroup$ Probably a bug or at least a limitation in calculus code. Limit will rely on Series and at least for elementary functions that will ignore branch cut issues unless assumptions are provided that give a clear indication of what side of such a cut we are on. $\endgroup$ Oct 16 '13 at 15:55
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It's not a bug in 11.1, but if this is still the case in 11.2, then it will be a bug. :)

Functions that take Assumptions are allowed to make assumptions when necessary. But they are also supposed to tell you when they do that. This is controlled by the GenerateConditions option. Limit in version 11.1 doesn't have this option. It's old, and I don't know that we have a defined behavior. But we're giving Limit a facelift in 11.2, including adding the option. When, unless you disable GenerateConditions, it should report the non-generic assumption x < 0. There will also be a dedicated function for taking limits of sequences instead of functions.

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  • $\begingroup$ How exciting! Since Limit is very related to Series, does that mean Series is getting a facelift too? $\endgroup$
    – QuantumDot
    Sep 1 '17 at 3:19
  • $\begingroup$ Not yet. I think it is safer to say that we will be laying down pieces for Series before we tackle it head on. $\endgroup$ Sep 4 '17 at 21:44
  • $\begingroup$ Cool; may I make a request? Could you make these functions so they are easily user-extensible? $\endgroup$
    – QuantumDot
    Sep 5 '17 at 1:44
  • $\begingroup$ You're certainly free to make the request... :) More seriously, I would suggest sending a report to support@wolfram.com with the suggestion, being as specific as possible. If it's logged in our database as something customers are asking for, it has a higher chance of being implemented. $\endgroup$ Sep 5 '17 at 18:54
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No problem with

Limit[(1/2 + ArcTan[n x]/Pi)^n, n -> Infinity, Assumptions -> x > 0]

$e^{-\frac{1}{\pi x}} $

Addition. In version 12.0

DiscreteLimit[(1/2 + ArcTan[n x]/Pi)^n, n -> Infinity,Assumptions -> x > 0]

$e^{-\frac{1}{\pi x}} $

DiscreteLimit[(1/2 + ArcTan[n x]/Pi)^n, n -> Infinity,Assumptions -> x <= 0]

$0$

Also

f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n
DiscreteLimit[f[x, n], n -> Infinity]
(*ConditionalExpression[E^(-(1/(\[Pi] x))), x > 0]*)
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Let's plot the limit of $f$ for a varying value of $x$ as $n$ approches $+\infty$:

DiscretePlot[Limit[f[x, n], n -> Infinity], {x, -10, 10, 0.5}]

enter image description here

Based on this discontinuity, I would not be surprised if Limit cannot handle it. Besides, when I evaluate Limit[f[x, n], n -> Infinity], I don't get a result. In conclusion, since you have two different regimes, you should specify which regime you are in.

Assuming[x < 0, Limit[f[x, n], n -> Infinity]]

Output: $0$

Assuming[x > 0, Limit[f[x, n], n -> Infinity]]

Output: $e^{-\frac{1}{\pi x}}$

These answer can be visually justified based on the following plots:

Plot[f[-1, n], {n, 0, 10}, Frame -> True, FrameLabel -> {"n", "f[1,n]"}]

enter image description here

Plot[f[+1, n], {n, 0, 10}, Frame -> True, FrameLabel -> {"n", "f[1,n]"}]

enter image description here

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    $Version
(* "12.1.1 for Mac OS X x86 (64-bit) (June 9, 2020)" *)

    f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n

    Limit[f[x, n], n -> Infinity]

Actual output

    InputForm[%]
ConditionalExpression[E^(-(1/(Pi*x))), x > 0]
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The intuitive meaning is that the large values of the Arctan function occur at lower x value as n gets bigger. The function series exhausts all the time the definition domain of the Arctan function. Taking the sum of Arctan function and 1/2 to the n-th power does not alter that.

So it is clear to make a division of the domain for convergences. Since the f[x,0] is the upper limit for all functions f==0 is the lower bound. The range between {0,1} suggests immediately that the series is converging equally to zero countered by potency.

I get with

$Version

12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)

Limit[f[x, n], n -> Infinity]

ConditionalExpression[E^(-(1/([Pi] x))), x > 0]

I can not reproduce the question's main topic.

The maths behind this is representing of the E^(-(1/(\[Pi] x))) function.

Plot[{tf, E^(-(1/(\[Pi] x)))}, {x, 0.001, 0.03}]

output

The convergence of the function series is equally on the domain of all reals.

This is useful to prove that

Limit[E^(-(1/(\[Pi] x))),x->0]=0

The function series contains special cases as x=1:

Table[ArcTan[n ], {n, -1, 1}]

{-([Pi]/4), 0, [Pi]/4}

Table[f[k, n], {k, -1, 1}]

{(1/2 - ArcTan[n]/[Pi])^n, 2^-n, (1/2 + ArcTan[n]/[Pi])^n}

while direct input into E^(-(1/([Pi] x))) results in 1/0 divide with singularity.

This is a valid graphical solution complete. The input and the Plot show everything that is necessary for mathematical terms for the solution: the behavior of the function, the common point is zero, strict monotony for positive real x and strictly falling with growing natural integer n and it shows to the bordering function exp(-pi/x) towards which the functions series convergences for n to infinity. It does not show the limit for large x or x=infinity, but that in the text of the answer. Everything is done.

There is are nice info from functions.wolfram.com and even better from Mathworld InverseTangent.

For example ArcTan[z]==-ArcTan[-z] for all complex z.

In formula 18 it offers an important identity:

ArcTan[z]=ArcCot[1/z]

And the follow the identities 22 to 25:

indentities

These generalize the question much further.

I stated I am not able to reproduce the question with my version of Mathematica. And supported the answer to the question that Mathematica gave me. I looked up some identities to make the answer of my Mathematica understandable instead of leaving it hidden behind Wolfram language input. I think that enhances understanding in a Wolfram Alpha fashion but using Mathworld.

x first, n first. Evaluations have to be interchangeable for convergence otherwise problems arise like suspected by this question which is not. Singular points are one source that changes behavior.

The question does not do the interchange that has to be central to prove reasons. So it stays behind. My concept of an answer does this needed switch of evaluation order. It lifts the level of the question up in order to reason about what happened inside the Limit-built-in. That is plain scientific attitude.

No. x is not fixed or unknown. x is the very same as n a variable. For each x the relation that fn[x,n] converges towards Exp[-1/(pi x)] for positive x has to be shown.

The methodology may start with x=0 and is trivial for each n and independent of n.

Then we have special cases where ArcTan.

Then make use of the properties of ArcTan:

(i) continuous, (ii) steadily increasing.

Then make use of the properties of each fn[x,n]

(i) continuous, (ii) steadily increasing, (iii) zero for x and all n, (iv) limited to all reals by zero, (v) limited for all reals be one.

Resolve[ForAll[
  x, (1/2 + ArcTan[(n + 1) x]/\[Pi])^(n + 1) - (1/2 + 
       ArcTan[n x]/\[Pi])^n > 0, x \[Element] Reals], {x, n}, Reals]

True

There is a possible more exact lower bound for positive reals that can be seen in graphic representations.

The use the last relation of the identities (22):

Plot[{Exp[-1/(\[Pi] x)], (1/2 + 
    ArcTan[ x]/\[Pi]), (1/2 + ArcTan[10 x]/\[Pi])^10, 
  Plot[{Exp[-1/(\[Pi] x)], (1/2 + 
      ArcTan[ x]/\[Pi]), (1-ArcTan[100/ x]/\[Pi])^100}, {x, .001, 
    1}]}, {x, .001, 1}]

output of plot

We have a new representation for our fn[x,n] without the constant 0.5 but still with the inverse potential of Pi.

Now there is the dependence on 1/x.

From the graphs of the fn[x,n] we see that there exists an integer for which all differences of fn+1-fn are closer together than a selected positive real espilon.

But what is out better limiting lower bound function like?

We have the dependence on 1/x so let's try testing function. We have the possibility to expand into the Taylor series and try to find a Taylor series for the limiting function. We may guess the limiting.

It can not be another transcendental function not even an inverse one.

We try Exp[-a/x] and fit the parameter a to the graphs and find comfortably a=1/Pi.

We get from

Limit[Limit[(ArcTan[1/(n y)]/\[Pi])^n, y -> x], n -> Infinity]

0

and

Limit[Limit[(ArcTan[1/(n y)]/\[Pi])^n, n -> Infinity], y -> x] 0

for all real x.

That is True but not what we already calculated.

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

ConditionalExpression[E^(-(1/([Pi] x))), x > 0]

Limit[(1 - ArcTan[1/(n x)]/\[Pi])^n, n -> Infinity]

E^(-(1/([Pi] x)))

Without the conditional expression. Both calculated in the same Mathematica.

So the simplified answer is that the identity reduces the domain for which the limiting function is the limiting function.

Might be that a representation for Exp[-1/(pi x)] as a limit of another function series does some more intimedation for the solution.

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    $\begingroup$ I'm not clear on whether or how this answers the question that was raised. $\endgroup$ Jun 21 '20 at 15:55
  • $\begingroup$ This is a valid graphical solution complete. The input and the Plot show everything that is necessary in mathematical terms for the solution: behavior of the function, common point is zero, strict monotony for positive real x and strictly falling with growing natural integer n and it show to the bordering function exp(-pi/x) towards which the functions series convergences for n to infinity. It does not show the limit for large x or x=infinity, but that in the text of the answer. Everything is done. $\endgroup$ Jun 22 '20 at 15:41
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    $\begingroup$ There was no question about whether the limit exists or what it is. The question asked was about why the limit given was placing an unstated condition on a parameter. A remark such as "The range between {0,1} suggests immediately that the series is converging equally to zero countered by potency." in no way helps to clarify matters. $\endgroup$ Jun 22 '20 at 18:07
  • $\begingroup$ I stated I am not able to reproduce the question with my version of Mathematica. And supported the answer for the question that Mathematica gave me. I looked up some identities to make the answer of my Mathematica understandable instead of leaving it hidden behind Wolfram language input. I think that enhances understanding in a Wolfram Alpha fashion but using Mathworld. x first, n first. Evaluations have to be interchangeable for convergence otherwise problems arise like suspected by this question which is not. Singular points are one source that changes behaviour. $\endgroup$ Jun 24 '20 at 8:39
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    $\begingroup$ I don't see an issue of interchanging operations in this case. The parameter x is fixed, albeit unknown (hence the need for multiple cases). If this were a multiple limit, that would be a (very) different matter, and order would of course matter. $\endgroup$ Jun 24 '20 at 13:39

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