Is 1^(infinity) indeterminate and can it be reduced to 0/0?

[uspatange]

Can anyone please tell me why 1^(infinity) is regarded to be indeterminate?
can it be reduced to 0/0? if yes, how?

 

[Tac-Tics]

Can anyone please tell me why 1^(infinity) is regarded to be indeterminate?
can it be reduced to 0/0? if yes, how?

Infinity is not a real number. There are no rules for how exponentiation works on something that isn't a number. It cannot be reduced to 0/0, because the expression is essentially meaningless, like asking the IQ of a banana.

 

[uspatange]

Infinity is not a real number. There are no rules for how exponentiation works on something that isn't a number. It cannot be reduced to 0/0, because the expression is essentially meaningless, like asking the IQ of a banana.

But still we can reduce forms like (infinity)-(infinity), (infinity)/(infinity), 0*(infinity) etc. to the form 0/0

 

[Tac-Tics]

But still we can reduce forms like (infinity)-(infinity), (infinity)/(infinity), 0*(infinity) etc. to the form 0/0

Not under the standard rules of algebra.

If you see infinity in an expression like that, you're probably seeing it as a shorthand for some kind of limit.

However, in the case of 1^\infty, we'd be dealing with something like \lim_{x \to \infty} 1^x which is perfectly reducible to 1, and *not* an indeterminate form.

And besides that, reducing an expression to 0/0 isn't helpful in the least.

 

[lurflurf]

1^(infinity)
is an indeterminate form because it is undefined (though as mentioned if the 1 is infact 1- we have 0 and if 1 is 1 we get 1 and only when 1 is 1+ doo we have an indeterminate form).
Suppose we are dealing with limits and not mere expresions in the (affinely) extended real numbers (R*). We can evaluate some limits using the (affinely) extended real numbers .
Example
lim_{x->infinity} sin(x)/x=0/infinity=0
works fine
lim_{x->0} sin(x)/x=0/0
does not because 0/0 is indeterminate ie undefined
this does not mean the limit cannot be found
we might use
1) Le'hopitals rule
2) Series expansion
3) algebraic manipulation

in particular when using Le'hopitals rule it is handy to exchange one form for another
ie
1^infinity->exp[ log( 1^infinity)]=exp[infinity*log(1)]->exp[log(1)/(1/infinity)]->exp[0/0]
example
lim_{x->infinity} (1+1/n)^n=exp[lim_{x->infinity} log(1+1/n)/(1/n)]
a 0/0 form
[log(1+1/n)]'=(-1/n^2)/(1+1/n)
(1/n)'=-1/n^2
so
exp[lim_{x->infinity} log(1+1/n)/(1/n)]=exp[lim_{x->infinity} [log(1+1/n)]'/[(1/n)]']
=exp[lim_{x->infinity} (-1/n^2)/(1+1/n)/(-1/n^2)]
=exp[lim_{x->infinity} 1/(1+1/n)]
=exp[ 1/(1+1/infinity)]
=exp[1/(1+0+)]
=exp[1/(1+)]
=exp[1-]
=e

 

[HallsofIvy]

Can anyone please tell me why 1^(infinity) is regarded to be indeterminate?
can it be reduced to 0/0? if yes, how?

"1/0" is referred to as "undefined" because the equation x= 1/0 is equivalent to 0*x= 1 which is false for all x: 0*x= 0 for all x, not 1. "0/0" is referred to as "indeterminant" because the equation "x= 0/0" is equivalent to 0*x= 0 which is true for all x. 1^infinity is "indeterminant" because x= 1^infinity is equivalent to ln(x)= infinity*ln(1)= \infinity*0= 0/(1/infinity)= 0/0.

And that's how you reduce it to 0/0. Of course, what you really reduce is an expression of the form g(x)^f(x) where g(x) goes to 1, f(x) goes to infinity. For example, (1/n)ⁿ, as n goes to infinity. Do what I suggested before: if y= ((n-1)/n)ⁿ= (1-1/n)ⁿ as n goes to infinity, then (1- 1/n) goes to 1 while n goes to infinity. Now ln(y)= n ln(1- 1/n)= ln(1- 1/n)/(1/n)= ln(1- m)/m as m goes to 0 the numerator goes to ln(1- 0)= ln(1)= 0 so this is of the form "0/0".

 

[stewartcs]

Can anyone please tell me why 1^(infinity) is regarded to be indeterminate?
can it be reduced to 0/0? if yes, how?

They are indeterminate since they do not have anyone particular solution. For example:

\lim_{x \to 0} \frac{x^2}{x} = 0

whereas

\lim_{x \to 0} \frac{x}{x} = 1

and

\lim_{x \to 0} \frac{x}{x^3} = \infty

Note that all of these are of the form 0/0.

Since the limit of something of the form 0/0 (in this case) does not have anyone particular value, it is deemed indeterminate.

With regard to indeterminate forms, one can reduce a function of the form 1^\infty to a function of the form 0/0.

Functions of the form 0/0 or \frac{\infty}{\infty} are useful when dealing with limits because one can directly apply L'Hopital's Rule.

Here's a link discussing it a bit more:

http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf

Hope this helps.

CS

 

[Tac-Tics]

1^infinity is "indeterminant" because x= 1^infinity is equivalent to ln(x)= infinity*ln(1)= \infinity*0= 0/(1/infinity)= 0/0.

So fast and loose with the algebra you are! But I suppose it works for what the OP is probably looking for.

Using that kind of reasoning, though, 1^\infty = 1 is just as accurate. For every real x, 1^x = 1. It's constant over all the reals, so by extension, it should be constant at both infinities as well.

To give a more accurate answer, it should be understood that division and exponentiation don't have anyone universally accepted definition. You can choose whatever definition you want. Some authors say that 0^0 = 1, and that definitions makes thing like Taylor series a little bit easier, because you can say exp(x) = \Sigma_{k=0} \frac{x^k}{k!} without having to work around the exponent... so we can achieve exp(0) = 0^0 + 0^1 + 0^2 + \cdots without having to make a fuss about the 0^0 term.

Why isn't 0^0 universally acknowledged to be equal to 1 then? Because then exponentiation wouldn't be continuous! Continuous functions are really nice, and we sort of wish that ^ would be continuous at all points of its domain. But there is no value for 0^0 which can make that so. The result... we simply say that (0, 0) is not in the domain of ^. It's a continuous function now at the cost that it is no longer defined on all of RxR.

When you talk about silly things like 1^\infty, we have similar issues. It's just the opposite situation, actually. We start with a continuous function (1^x) with a restricted domain (all real numbers). By extending the domain to include infinities, we lose continuity.

But I'm just picky about those sorts of things =-) If 1^\infty = \frac{0}{0} seems like it should be true, go for it.

 

[adriank]

I'm pretty sure what we're dealing here with limits of the form 1^\infty is that the 1 is allowed to vary as well, so we're considering the limit \displaystyle\lim_{(a, b)\to(1, \infty)} a^b, which doesn't exist: If t \to \infty, then 1 + 1/t \to 1, t \to \infty, and 1^t \to 1 and (1+1/t)^t \to e, so 1^\infty is indeterminate.