There are ways to protect the path leading to `1**oo` but the expression 
itself is agnostic to path. An appeal that it give 1 has been made here: 
https://github.com/sympy/sympy/pull/21619

/c

On Saturday, November 6, 2021 at 11:55:14 AM UTC-5 [email protected] 
wrote:

> I am just a hobby mathematician, but it seems to me like this:
>
> 1^oo := lim(1^n) = lim(1) = 1.
>
> The other 'limits' seem to me to be an inadmissible 'exchange' of limits:
> 1 != (1 + 1/n) for any finite n 
>
> On Sat 6. Nov 2021 at 14:15, Oscar Benjamin <[email protected]> wrote:
>
>> On Sat, 6 Nov 2021 at 11:58, Anderson Bhat <[email protected]> wrote:
>> >
>> > Hello guys , I am working on couple of Pr's extending the functionality 
>> of the doit method in the concrete module , I noticed that one 
>> inconsistency leads to couple of errors . Product(1, (n, 1, oo)).doit() 
>> returns 1 and 1**oo returns NaN. Other integers work as expected . These 
>> expressions are equivalent right ??? or am I missing something !
>>
>> The expression 1**oo is indeterminate because there are different ways
>> that you could arrive at this form that have different limits:
>>
>> In [31]: limit((1 + 1/n)**n, n, oo)
>> Out[31]: ℯ
>>
>> In [32]: limit((1 + 1/n**2)**n, n, oo)
>> Out[32]: 1
>>
>> In [33]: limit((1 + 1/sqrt(n))**n, n, oo)
>> Out[33]: ∞
>>
>> The product in your case defines a particular limit so it is not 
>> indeterminate:
>>
>> In [38]: Product(1, (n, 1, m)).doit()
>> Out[38]: 1
>>
>> In [39]: limit(Product(1, (n, 1, m)).doit(), m, oo)
>> Out[39]: 1
>>
>> --
>> Oscar
>>
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>> .
>>
> -- 
> Best regards,
>
> Peter Stahlecker
>

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