Let \[y= 1^n, \lim_{n\to\infty}y=\lim_{n\to\infty}1^n, \lim_{n\to\infty} \ln y=\lim_{n\to\infty} \ln1^n=\lim_{n\to\infty}n \ln 1=\lim_{n\to\infty}\frac{\ln 1}{\frac 1n} =\frac00\]
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Lab Bhattacharjee
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3 years, 10 months ago

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@Lab Bhattacharjee
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No, \(\ln 1\) is always 0 but \(\frac{1}{n}\) is just approach 0 so \(\lim_{n->\infty}\frac{\ln 1}{1/n} = \lim_{n->\infty}\frac{0}{1/n} = 0\) .So \(y=1\). The problem about this indetermination is that the 1 isn't exactly 1, it is just a symbol for some function that approach 1, like \(1-1/x\), so that \(1^\infty\) is indeterminate
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Pitchayatak Ponrod
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3 years, 10 months ago

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@Lab Bhattacharjee
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Note, the last step is not valid. You have not justified that \( \lim \frac{f(x)}{g(x)} = \frac { \lim f(x)} { \lim g(x) } \), which is only true under certain conditions which your functions do not satisfy.

For example, \( \frac{1}{2} = \lim_{n \rightarrow \infty} \frac{n}{2n} \), but you cannot split it apart after that.

Review L'Hopital Rule.
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Calvin Lin
Staff
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3 years, 10 months ago

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I think what you mean is really \(\lim_{x\rightarrow\infty} 1^x\)
Furthermore, this might be better stated as \(\lim_{t\rightarrow1}\lim_{x\rightarrow\infty}t^x\)
If we let the function inside the outer limit be \(f(t)\), then clearly \(f(t)=0\) for \(-1<t<1\) and \(f(t)\) is \(\infty\) for \(t>1\). This leads me to tentatively conclude that the limit is undefined, although it is defined for either direction.
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Matthew Lipman
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3 years, 10 months ago

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I disprove that, let's see this.

\(1^{-2} = 1\)

\(1^{-1} = 1\)

\(1^{-\frac{1}{2}} = 1\)

\(1^{0} = 1\)

\(1^{\frac{1}{3}} = 1\)

\(1^{1} = 1\)

\(1^{100} = 1\)

so, \(1^{\infty} = 1\)

because \(1\) act as a unit to count everything (\(\infty\)), it make everything incalculable (\(\infty\)) become calculable. so, the \(1\) is basic of everything. it means everything has \(1\) as its factor or its characteristic, even all negatives have \(1\) as their factor too.

we must determine the range value of \(1\) first before we able to calculate everything.

Despite what anyone else says about infinity being a concept and not an actual number, I think this is an interesting topic and I view the resentment people have towards using infinity as having an algebraic use as a limiting mindset (just like not accepting imaginary numbers as numbers was for those thousands of years). I saw several posts saying that you should take a limit instead but the problem I see here is that this will give a real number which is not the same as infinity. Say you're trying to find (-1)^infinity and use a limit will give -1 or 1 depending on whether the limit comes out even or odd, yet I can argue that (-1)^infinity will be zero. And I do agree that 1^infinity is indeterminate. I have some papers stashed away somewhere in which I came to some interesting conclusions regarding raising number to infinity that I'll try to find and share.
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Hyrum Hammon
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3 years, 10 months ago

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Rigorously, what you want to show is that if lim f(t)=1 and lim g(t)=infinity for some t and functions f and g, then lim f(t)^g(t) is not necessarily any particular value. This is simple to show. Consider, for some positive number a, the functions f(x)=a^(1/x) and g(x)=x setting our t=0. Then lim(t->0) f(t)^g(t)=lim (a^(1/x))^x=lim a^(x/x) = a, showing that the limit could be any positive real number. It could also be infinity, considering f(x)=e^(1/(x^2)) and g(x)=x.
Intuitively, 1^infinity is indeterminate because the rates at which something goes to 1 and something else goes to infinity matter. It can also be explained by the fact that log(1^infinity)=infinitylog(1)= infinity0, a well known indeterminate.
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Jacob Gurev
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3 years, 10 months ago

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I think it should tend to 1 as 1 raised to any large number like 9999999999 is 1 itself, so is infinity I guess. Note, this is true only when 1 is exactly 1. Even if it fluctuates by .0001 this limit won't tend to 1. So, you can also say its inderterminate. I think the answer depends on the situation in which you are. Imagine you need to explain this for some measurement. A measurement is never fully accurate. For example you cannot balance a copy on a pin because you can never accurately mark its center of mass. It will eventually fall down. In the same way marking the situation in which 1 is not exactly 1, we can't say it tends to 1. Instead, we ought to say its indeterminant form. This is my own theory from my own experience.
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Lokesh Sharma
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3 years, 10 months ago

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Anyone tried making graph of 1^x ? Doesn't looks like indeterminate form when x-->infinity. It seems to be exactly 1.

On the other hand lim (x-->inf, y-->1) y^x doesn't exists (or is indeterminate) because Left Hand Limit (=0) is not equal to Right Hand Limit (tends to infinity).
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Mirza Baig
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3 years, 10 months ago

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Here follows an intuitive idea. Sorry I didn't check posts below to see if someone has posted something similar. But \(\lim_{n\to \infty}\lim_{k\to 1^{+}} k^{n}\) is infinity since it is greater than 1 it shoots up. Again, \(\lim_{n\to \infty}\lim_{k\to 1^{-}} k^{n}=0\) since its less than 1 it plummets.
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Abhishek De
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3 years, 10 months ago

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Consider this:
We can write 1^infinity as:
\[(1+0)^{1/0}\]
This can be written as:
\[\lim_{x\to0}(1+x)^{1/x}\]
This limit is equal to \[e=2.718\]
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Shubham Srivastava
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3 years, 10 months ago

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@Shubham Srivastava
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This is a good idea, but unfortunately, there are some problems. For example, 0=2*0, so we could replace the x in the limit with 2x, which changes the result, etc. The problem arises b/c infinity and 1/0 are distinct concepts.
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Matthew Lipman
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3 years, 10 months ago

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@Shubham Srivastava
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Your first statement isn't true, and Matthew subtly touched on the reason. You cannot simply say that \( \lim f(x) ^{g(x)} = \left( \lim f(x) \right) ^{ \lim g(x) } \), but have to justify it. This is similar to saying that \( \lim \frac{ f(x)} {g(x) } = \frac { \lim f(x) } { \lim g(x) } \); you need to check the conditions under which this statement is true.
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Calvin Lin
Staff
·
3 years, 10 months ago

INFINITY is actually a dimensional extension!! it is not a number!!! it is actually a conception that space can be situated in INFINITY!!!
–
Subhrodipto Basu Choudhury
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3 years, 10 months ago

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here 1 is something like 1.0000000000000000000000000...................000001 or 1.99999999999999999999999999999999999...............99999999999 , therefore these numbers raised to the power infinity are meaningless and thus it is indeterminate form.
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Harsh Mohan
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3 years, 10 months ago

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@Harsh Mohan
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1^infinity is 1 but (->1)^infinity is indeterminate....
understand the difference between 1 and (->1),

(->1)^infinity is indeterminate because it depends on what type of function approaching one is raised to infinity
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Jatin Yadav
·
3 years, 10 months ago

## Comments

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TopNewestinfinite isn't a number,a common misconceptionn – Beakal Tiliksew · 3 years, 10 months ago

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– Jeffrey Robles · 3 years, 10 months ago

Yeah. It is a misleading question in the first place, I guess. Thanks!Log in to reply

Let \[y= 1^n, \lim_{n\to\infty}y=\lim_{n\to\infty}1^n, \lim_{n\to\infty} \ln y=\lim_{n\to\infty} \ln1^n=\lim_{n\to\infty}n \ln 1=\lim_{n\to\infty}\frac{\ln 1}{\frac 1n} =\frac00\] – Lab Bhattacharjee · 3 years, 10 months ago

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– Pitchayatak Ponrod · 3 years, 10 months ago

No, \(\ln 1\) is always 0 but \(\frac{1}{n}\) is just approach 0 so \(\lim_{n->\infty}\frac{\ln 1}{1/n} = \lim_{n->\infty}\frac{0}{1/n} = 0\) .So \(y=1\). The problem about this indetermination is that the 1 isn't exactly 1, it is just a symbol for some function that approach 1, like \(1-1/x\), so that \(1^\infty\) is indeterminateLog in to reply

For example, \( \frac{1}{2} = \lim_{n \rightarrow \infty} \frac{n}{2n} \), but you cannot split it apart after that.

Review L'Hopital Rule. – Calvin Lin Staff · 3 years, 10 months ago

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I think what you mean is really \(\lim_{x\rightarrow\infty} 1^x\) Furthermore, this might be better stated as \(\lim_{t\rightarrow1}\lim_{x\rightarrow\infty}t^x\) If we let the function inside the outer limit be \(f(t)\), then clearly \(f(t)=0\) for \(-1<t<1\) and \(f(t)\) is \(\infty\) for \(t>1\). This leads me to tentatively conclude that the limit is undefined, although it is defined for either direction. – Matthew Lipman · 3 years, 10 months ago

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I disprove that, let's see this.

\(1^{-2} = 1\)

\(1^{-1} = 1\)

\(1^{-\frac{1}{2}} = 1\)

\(1^{0} = 1\)

\(1^{\frac{1}{3}} = 1\)

\(1^{1} = 1\)

\(1^{100} = 1\)

so, \(1^{\infty} = 1\)

because \(1\) act as a unit to count everything (\(\infty\)), it make everything incalculable (\(\infty\)) become calculable. so, the \(1\) is basic of everything. it means everything has \(1\) as its factor or its characteristic, even all negatives have \(1\) as their factor too.

we must determine the range value of \(1\) first before we able to calculate everything.

YIS – Yulianto Indra Setiawan · 3 years, 10 months ago

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Despite what anyone else says about infinity being a concept and not an actual number, I think this is an interesting topic and I view the resentment people have towards using infinity as having an algebraic use as a limiting mindset (just like not accepting imaginary numbers as numbers was for those thousands of years). I saw several posts saying that you should take a limit instead but the problem I see here is that this will give a real number which is not the same as infinity. Say you're trying to find (-1)^infinity and use a limit will give -1 or 1 depending on whether the limit comes out even or odd, yet I can argue that (-1)^infinity will be zero. And I do agree that 1^infinity is indeterminate. I have some papers stashed away somewhere in which I came to some interesting conclusions regarding raising number to infinity that I'll try to find and share. – Hyrum Hammon · 3 years, 10 months ago

Log in to reply

Rigorously, what you want to show is that if lim f(t)=1 and lim g(t)=infinity for some t and functions f and g, then lim f(t)^g(t) is not necessarily any particular value. This is simple to show. Consider, for some positive number a, the functions f(x)=a^(1/x) and g(x)=x setting our t=0. Then lim(t->0) f(t)^g(t)=lim (a^(1/x))^x=lim a^(x/x) = a, showing that the limit could be any positive real number. It could also be infinity, considering f(x)=e^(1/(x^2)) and g(x)=x. Intuitively, 1^infinity is indeterminate because the rates at which something goes to 1 and something else goes to infinity matter. It can also be explained by the fact that log(1^infinity)=infinity

log(1)= infinity0, a well known indeterminate. – Jacob Gurev · 3 years, 10 months agoLog in to reply

I think it should tend to 1 as 1 raised to any large number like 9999999999 is 1 itself, so is infinity I guess. Note, this is true only when 1 is exactly 1. Even if it fluctuates by .0001 this limit won't tend to 1. So, you can also say its inderterminate. I think the answer depends on the situation in which you are. Imagine you need to explain this for some measurement. A measurement is never fully accurate. For example you cannot balance a copy on a pin because you can never accurately mark its center of mass. It will eventually fall down. In the same way marking the situation in which 1 is not exactly 1, we can't say it tends to 1. Instead, we ought to say its indeterminant form. This is my own theory from my own experience. – Lokesh Sharma · 3 years, 10 months ago

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Anyone tried making graph of 1^x ? Doesn't looks like indeterminate form when x-->infinity. It seems to be exactly 1.

On the other hand lim (x-->inf, y-->1) y^x doesn't exists (or is indeterminate) because Left Hand Limit (=0) is not equal to Right Hand Limit (tends to infinity). – Mirza Baig · 3 years, 10 months ago

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Here follows an intuitive idea. Sorry I didn't check posts below to see if someone has posted something similar. But \(\lim_{n\to \infty}\lim_{k\to 1^{+}} k^{n}\) is infinity since it is greater than 1 it shoots up. Again, \(\lim_{n\to \infty}\lim_{k\to 1^{-}} k^{n}=0\) since its less than 1 it plummets. – Abhishek De · 3 years, 10 months ago

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Consider this: We can write 1^infinity as: \[(1+0)^{1/0}\] This can be written as: \[\lim_{x\to0}(1+x)^{1/x}\] This

limitis equal to \[e=2.718\] – Shubham Srivastava · 3 years, 10 months agoLog in to reply

– Matthew Lipman · 3 years, 10 months ago

This is a good idea, but unfortunately, there are some problems. For example, 0=2*0, so we could replace the x in the limit with 2x, which changes the result, etc. The problem arises b/c infinity and 1/0 are distinct concepts.Log in to reply

– Calvin Lin Staff · 3 years, 10 months ago

Your first statement isn't true, and Matthew subtly touched on the reason. You cannot simply say that \( \lim f(x) ^{g(x)} = \left( \lim f(x) \right) ^{ \lim g(x) } \), but have to justify it. This is similar to saying that \( \lim \frac{ f(x)} {g(x) } = \frac { \lim f(x) } { \lim g(x) } \); you need to check the conditions under which this statement is true.Log in to reply

– Vikrant Desai · 3 years, 10 months ago

so intelligent...:)Log in to reply

INFINITY is actually a dimensional extension!! it is not a number!!! it is actually a conception that space can be situated in INFINITY!!! – Subhrodipto Basu Choudhury · 3 years, 10 months ago

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here 1 is something like 1.0000000000000000000000000...................000001 or 1.99999999999999999999999999999999999...............99999999999 , therefore these numbers raised to the power infinity are meaningless and thus it is indeterminate form. – Harsh Mohan · 3 years, 10 months ago

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(->1)^infinity is indeterminate because it depends on what type of function approaching one is raised to infinity – Jatin Yadav · 3 years, 10 months ago

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– Jatin Yadav · 3 years, 10 months ago

check the above@ Calvin, ain't it right???Log in to reply