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

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.

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.

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.

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.

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.

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).

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.

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.

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.

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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TopNewestinfinite isn't a number,a common misconceptionn

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

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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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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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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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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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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

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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.

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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)=infinity

log(1)= infinity0, a well known indeterminate.Log 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.

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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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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 \[e=2.718\]Log in to reply

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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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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so intelligent...:)

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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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INFINITY is actually a dimensional extension!! it is not a number!!! it is actually a conception that space can be situated in INFINITY!!!

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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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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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check the above@ Calvin, ain't it right???

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