Indeterminate Forms

Debsta

MHF Helper
Oct 2009
1,313
605
Brisbane
Jhevon, I understand what you are saying about conflicting ideas and that is certainly why 0^0 is indeterminate.

But the limit of a^x as x-> infinity depends on what a is.

If a >1 then the limit doesn't exist.
If 0<a<1, then the limit is 0.
If a=0, then the limit is 0.
If a =1, then the limit is 1.
If a<0 , then all hell breaks loose and so no limit exists.

Jhevon, you 've said that "1^infinity is indeterminate because the base of 1 should give you 1, but thinking of exponential functions, they go off to infinity as the power goes to infinity." But, as you later said, that only applies if the base is greater than 1. So there's no contradiction there as there is for 0^0.0

I don't agree that the limit of 1^x as x-> infinity is indeterminate. Even though that's what wikipedia says - who believes them anyway!

1^infinity = 1x1x1x1x...... How can it not be 1? Where does it change from being 1 to being indeterminate?

Topsquark, I find it interesting that you learnt that it is indeterminate and not 1.

I'm open to anyone convincing me otherwise. Interesting thread!!
 
  • Like
Reactions: topsquark

Jhevon

MHF Helper
Feb 2007
11,681
4,225
New York, USA
Jhevon, I understand what you are saying about conflicting ideas and that is certainly why 0^0 is indeterminate.

But the limit of a^x as x-> infinity depends on what a is.

If a >1 then the limit doesn't exist.
If 0<a<1, then the limit is 0.
If a=0, then the limit is 0.
If a =1, then the limit is 1.
If a<0 , then all hell breaks loose and so no limit exists.

Jhevon, you 've said that "1^infinity is indeterminate because the base of 1 should give you 1, but thinking of exponential functions, they go off to infinity as the power goes to infinity." But, as you later said, that only applies if the base is greater than 1. So there's no contradiction there as there is for 0^0.0

I don't agree that the limit of 1^x as x-> infinity is indeterminate. Even though that's what wikipedia says - who believes them anyway!

1^infinity = 1x1x1x1x...... How can it not be 1? Where does it change from being 1 to being indeterminate?

Topsquark, I find it interesting that you learnt that it is indeterminate and not 1.

I'm open to anyone convincing me otherwise. Interesting thread!!
Have you seen the definition of the number e? The base approaches 1 while the power approaches infinity. I believe some examples of this were given in an earlier post. A limit form approaching \(\displaystyle 1^\infty\) therefore cannot be determinate. We know of specific examples where the answer has variations, it may be 1, it also may not be and it may also diverge.

Read the part of my thread about the nuances.

I think I know where your confusion lies: the base is APPROACHING 1 in these forms, it's not identically equal to 1. We're talking about limit forms, not functions.
 
  • Like
Reactions: topsquark

Debsta

MHF Helper
Oct 2009
1,313
605
Brisbane
Yeah I'm familiar with the limit definition of e. But in the case of 1^infinity, the base isn't approaching 1, it IS 1.

I think your bit about nuances, negated your argument above ìt. a^x only approaches infinity as x approaches infinity, IFF a>1. It doesn't say anything about what happens when a=1.
 

Jhevon

MHF Helper
Feb 2007
11,681
4,225
New York, USA
Yeah I'm familiar with the limit definition of e. But in the case of 1^infinity, the base isn't approaching 1, it IS 1.

I think your bit about nuances, negated your argument above ìt. a^x only approaches infinity as x approaches infinity, IFF a>1. It doesn't say anything about what happens when a=1.
When we write \(\displaystyle 1^\infty\), it is understood to be shorthand. You can't actually plug in infinity as an exponent. The indeterminate form \(\displaystyle 1^\infty\) means, there are variable expressions \(\displaystyle A\) and \(\displaystyle B\) such that \(\displaystyle \lim_{x\to a}A = 1\) and \(\displaystyle \lim_{x \to a}B = \infty\) and we are considering the limit \(\displaystyle \lim_{x \to a}A^B\). This is indeterminate, and we express this form as \(\displaystyle 1^\infty\).

As I mentioned, the "1" in the form is not identically 1. It is indeed true that \(\displaystyle \lim_{x \to \infty} 1^x = 1\), because that is not an indeterminate form. \(\displaystyle 1^x\) is identically 1 for any real number \(\displaystyle x\). The same cannot be said for \(\displaystyle \left( 1 + \frac 1x \right)^x\), for example, hence the introduction of a competing idea.

Now, if you look at what I said in light of this, I think you will see the nuance I am talking about. Normally when we write \(\displaystyle \lim_{x \to \infty} a^x\), it is assumed that \(\displaystyle a\) is identically some constant. The moment that assumption is removed, and you allow \(\displaystyle a\) to depend on \(\displaystyle x\), problems arise. In particular, here, problems arise if the expression for \(\displaystyle a(x)\) goes to 1. On one hand the answer *was* clear (you yourself laid out the answers to all cases based on the assumed value of \(\displaystyle a\)). On the other hand, if we allow \(\displaystyle a\) to depend on \(\displaystyle x\) in a particular way, an idea is introduced that made it unclear--hence, indeterminate form.
 
Last edited:

Debsta

MHF Helper
Oct 2009
1,313
605
Brisbane
You said
1575205217721.png
I have never come across that shorthand before. I would read \(\displaystyle 1^\infty\) as 1575205370021.png.

I think that is where the confusion lies (I realise you can't actually plug in infinity as an exponent).

You also said that
1575205008387.png
That is what I have said all along, there's no confusion there.
 

Attachments

  • Like
Reactions: Jhevon

Jhevon

MHF Helper
Feb 2007
11,681
4,225
New York, USA
Yes, \(\displaystyle 1^\infty\) is interpreted how I wrote it. When someone writes it down as an indeterminate form, they mean that the base approaches 1 (like in the definition of \(\displaystyle e\), or \(\displaystyle \frac 1e\)). There are a few occasions where one would interpret such notation as \(\displaystyle \lim_{x \to \infty}1^x\), like is one is getting lazy in writing out an improper integral, I've also seen similar notation in a measure theory class (where a power series background was assumed), but also because one was getting lazy to write out full notation. @greg1313 posted several examples to illustrate what we mean by the indeterminate form \(\displaystyle 1^\infty\), @Idea's post was also insightful in that regard.

It's funny how sometimes people can end up disagreeing and it's all based on some mis-notation or interpretation of notation!

Proper notation is important kids.
 
Last edited:
  • Like
Reactions: Debsta

Jhevon

MHF Helper
Feb 2007
11,681
4,225
New York, USA
This is mostly for @orcsmoocher: In light of the above misunderstanding, I suppose it is worth mentioning that that is how all these forms should be interpreted. So in the original list, where you see \(\displaystyle 1 \cdot \infty\) for example. That should be interpreted as we have variable expressions \(\displaystyle A,B\) depending on \(\displaystyle x\) such that \(\displaystyle \lim_{x \to a}A = 1\) and \(\displaystyle \lim_{x \to a}B = \infty\) and we are considering a limit of the form \(\displaystyle \lim_{x \to a}A \cdot B\). So don't think of the constants here as being identically those constants. That may have also been the interpretation of the OP.

(Now, come to think of it, I assigned a similar problem to my calculus 2 class this semester, and some students got frustrated enough to randomly change all the options until they got it right. I wonder if the notation was not understood...hmmm.)

* Note that in the above limits, the \(\displaystyle a\) may represent \(\displaystyle + \infty\) or \(\displaystyle - \infty\).
 
Nov 2019
23
0
Depths of Hell, babeeey
This is mostly for @orcsmoocher: In light of the above misunderstanding, I suppose it is worth mentioning that that is how all these forms should be interpreted. So in the original list, where you see \(\displaystyle 1 \cdot \infty\) for example. That should be interpreted as we have variable expressions \(\displaystyle A,B\) depending on \(\displaystyle x\) such that \(\displaystyle \lim_{x \to a}A = 1\) and \(\displaystyle \lim_{x \to a}B = \infty\) and we are considering a limit of the form \(\displaystyle \lim_{x \to a}A \cdot B\). So don't think of the constants here as being identically those constants. That may have also been the interpretation of the OP.

(Now, come to think of it, I assigned a similar problem to my calculus 2 class this semester, and some students got frustrated enough to randomly change all the options until they got it right. I wonder if the notation was not understood...hmmm.)

* Note that in the above limits, the \(\displaystyle a\) may represent \(\displaystyle + \infty\) or \(\displaystyle - \infty\).
Thank you all for the help! 😊 I think I understand it now, and I'll do some more practice after my next workset.
 

Debsta

MHF Helper
Oct 2009
1,313
605
Brisbane
Yes, \(\displaystyle 1^\infty\) is interpreted how I wrote it. When someone writes it down as an indeterminate form, they mean that the base approaches 1 (like in the definition of \(\displaystyle e\), or \(\displaystyle \frac 1e\)). There are a few occasions where one would interpret such notation as \(\displaystyle \lim_{x \to \infty}1^x\), like is one is getting lazy in writing out an improper integral, I've also seen similar notation in a measure theory class (where a power series background was assumed), but also because one was getting lazy to write out full notation. @greg1313 posted several examples to illustrate what we mean by the indeterminate form \(\displaystyle 1^\infty\), @Idea's post was also insightful in that regard.

It's funny how sometimes people can end up disagreeing and it's all based on some mis-notation or interpretation of notation!

Proper notation is important kids.
Thanks Jhevon. It's clear now that I was reading the notation incorrectly. I've just never seen that interpretation before. Don't know why I've never come across it before! Must have skipped that lecture! o_O