One to the power of infinity: I don't agree with her that 1^∞ is one of the indereminant forms

[mario99]

I am a professional calculus student while my friend's friend has just taken calculus I.

She says that [imath]1^\infty[/imath] is one of the indeterminant forms and there is a special technique to solve it.

I told her, proceed please. She says:

Let

[imath]\displaystyle y = 1^{\infty} = \lim_{n\rightarrow \infty} 1^n [/imath]

[imath]\displaystyle e^{\ln y} = \lim_{n\rightarrow \infty} e^{\ln 1^n} [/imath]

[imath]\displaystyle y = \lim_{n\rightarrow \infty} e^{n\ln 1} [/imath]

[imath]y = e^{\lim_{n\rightarrow \infty}n\ln 1} = e^{\lim_{n\rightarrow \infty} 0} = e^0 = 1[/imath]

While her technique was not bad to solve limits to the power of [imath]n[/imath], I don't agree with her that [imath]1^{\infty}[/imath] is one of the indereminant forms because we can always safely say [imath]1^{\infty} = 1[/imath].

From my long noob experience, I consider indeterminant form as a machine that will give us different numbers depending on the original argument of the limit where [imath]1^{\infty}[/imath] is not the case here. Where did I go wrong with my conclusion?

 

[Dr.Peterson]

I am a professional calculus student while my friend's friend has just taken calculus I.

She says that [imath]1^\infty[/imath] is one of the indeterminant forms and there is a special technique to solve it.

I told her, proceed please. She says:

Let

[imath]\displaystyle y = 1^{\infty} = \lim_{n\rightarrow \infty} 1^n [/imath]

[imath]\displaystyle e^{\ln y} = \lim_{n\rightarrow \infty} e^{\ln 1^n} [/imath]

[imath]\displaystyle y = \lim_{n\rightarrow \infty} e^{n\ln 1} [/imath]

[imath]y = e^{\lim_{n\rightarrow \infty}n\ln 1} = e^{\lim_{n\rightarrow \infty} 0} = e^0 = 1[/imath]

While her technique was not bad to solve limits to the power of [imath]n[/imath], I don't agree with her that [imath]1^{\infty}[/imath] is one of the indereminant forms because we can always safely say [imath]1^{\infty} = 1[/imath].

From my long noob experience, I consider indeterminant form as a machine that will give us different numbers depending on the original argument of the limit where [imath]1^{\infty}[/imath] is not the case here. Where did I go wrong with my conclusion?

I'm a little confused. You say that she called it indeterminate, yet found it to be equal to 1, while you say it is not indeterminate and that it is always equal to 1. Who is saying what? How does what she says support her claim? How do you support yours, in contrast?

But in fact [imath]1^\infty[/imath] is an indeterminate form; it does not always equal 1.

Indeterminate forms involve limits combining two functions, which depending on how they approach their limits, can produce different combined limits. In this case, a limit of f^g, where f approaches 1 and g approaches infinity, can take on different values. In one case, you can get 1, but in other cases you can get other limits. Being able to find the limit in one case doesn't mean that it isn't indeterminate.

The mistake in her work, if she were trying to show it is indeterminate, is in taking the 1 as a constant. Since 1 to any power is 1, that forces the answer to be 1. That form is not indeterminate.

What you need to do is to replace the base, 1, with a function that approaches 1, and try to get a limit other than 1. It can be done! (Hint: one nice choice of functions gives the limit e.)

 

[mario99]

Thank you Dr.Peterson for guiding me.

I'm a little confused. You say that she called it indeterminate, yet found it to be equal to 1, while you say it is not indeterminate and that it is always equal to 1. Who is saying what? How does what she says support her claim? How do you support yours, in contrast?

I, Mario, say [imath]1^{\infty}[/imath] is not an indeterminate form, but she says the reverse.

Indeterminate forms involve limits combining two functions, which depending on how they approach their limits, can produce different

Here I agree with you. If two functions or more, led to the form [imath]1^{\infty}[/imath], I would say it is in indeterminant form, and its solution not necessary to be [imath]1[/imath]. But if [imath]1^{n}[/imath] was given as the argument of the limit, it would not be considered as an indeterminant form and that was my claim.

She was wrong, and I was the winner, weren't I?

 

[Dr.Peterson]

I, Mario, say [imath]1^{\infty}[/imath] is not an indeterminate form, but she says the reverse.
She was wrong, and I was the winner, weren't I?

No, it is an indeterminate form. The emphasis has to be on what we mean by "indeterminate form". It does not mean that the 1 is a constant; it means that the base is a function approaching 1. If it meant the latter, you would be right; but that is not what people mean when they say this.

Surely you know that this is a standard fact:

Indeterminate form - Wikipedia

en.wikipedia.org

For example, it is unclear what the following expressions ought to evaluate to:[1]​

​

These seven expressions are known as indeterminate forms. More specifically, such expressions are obtained by naively applying the algebraic limit theorem to evaluate the limit of the corresponding arithmetic operation of two functions, yet there are examples of pairs of functions that after being operated on converge to 0, converge to another finite value, diverge to infinity or just diverge. This inability to decide what the limit ought to be explains why these forms are regarded as indeterminate. A limit confirmed to be infinity is not indeterminate since it has been determined to have a specific value (infinity).[1] The term was originally introduced by Cauchy's student Moigno in the middle of the 19th century.​

I presume you are not being serious.

Here I agree with you. If two functions or more, led to the form [imath]1^{\infty}[/imath], I would say it is in indeterminant form, and its solution not necessary to be [imath]1[/imath]. But if [imath]1^{n}[/imath] was given as the argument of the limit, it would not be considered as an indeterminant form and that was my claim.

First, the word is "indeterminate". (And the form relates to exactly two functions, not more.)

Second, as I said above, what you say here is simply not what we mean when we talk about that as a form. What you are saying is, at best, misleading.

And I still can't tell what her point was in giving the example. It's the wrong example if she was trying to show that it is indeterminate. If she was just trying to show that indeterminate forms do not mean that the limit can't be evaluated, that's true

 

[mario99]

No, it is an indeterminate form. The emphasis has to be on what we mean by "indeterminate form". It does not mean that the 1 is a constant; it means that the base is a function approaching 1. If it meant the latter, you would be right; but that is not what people mean when they say this.

Surely you know that this is a standard fact:

Indeterminate form - Wikipedia

en.wikipedia.org

For example, it is unclear what the following expressions ought to evaluate to:[1]​

​

These seven expressions are known as indeterminate forms. More specifically, such expressions are obtained by naively applying the algebraic limit theorem to evaluate the limit of the corresponding arithmetic operation of two functions, yet there are examples of pairs of functions that after being operated on converge to 0, converge to another finite value, diverge to infinity or just diverge. This inability to decide what the limit ought to be explains why these forms are regarded as indeterminate. A limit confirmed to be infinity is not indeterminate since it has been determined to have a specific value (infinity).[1] The term was originally introduced by Cauchy's student Moigno in the middle of the 19th century.​

I presume you are not being serious.


First, the word is "indeterminate". (And the form relates to exactly two functions, not more.)

Second, as I said above, what you say here is simply not what we mean when we talk about that as a form. What you are saying is, at best, misleading.

And I still can't tell what her point was in giving the example. It's the wrong example if she was trying to show that it is indeterminate. If she was just trying to show that indeterminate forms do not mean that the limit can't be evaluated, that's true

Thank you a lot for the detailed explanation Dr.Peterson. I had a discussion with her later and she claimed that she meant two function approaching [imath]1^\infty[/imath] which is in this case indereminate. But I am sure she is lying. She understood the concept falsely. It is later she corrected it. I consider myself winning.