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# 1^infinity

How do we prove/disprove that 1^inifinity is indeterminate?

Note by Jeffrey Robles
3 years, 8 months ago

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infinite isn't a number,a common misconceptionn · 3 years, 8 months ago

Yeah. It is a misleading question in the first place, I guess. Thanks! · 3 years, 8 months ago

Let $y= 1^n, \lim_{n\to\infty}y=\lim_{n\to\infty}1^n, \lim_{n\to\infty} \ln y=\lim_{n\to\infty} \ln1^n=\lim_{n\to\infty}n \ln 1=\lim_{n\to\infty}\frac{\ln 1}{\frac 1n} =\frac00$ · 3 years, 8 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 indeterminate · 3 years, 8 months ago

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. Staff · 3 years, 8 months ago

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. · 3 years, 8 months ago

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 · 3 years, 8 months ago

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. · 3 years, 8 months ago

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. · 3 years, 8 months ago

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. · 3 years, 8 months ago

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). · 3 years, 8 months ago

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. · 3 years, 8 months ago

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 $e=2.718$ · 3 years, 8 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. · 3 years, 8 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. Staff · 3 years, 8 months ago

so intelligent...:) · 3 years, 8 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!!! · 3 years, 8 months ago

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. · 3 years, 8 months ago

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 · 3 years, 8 months ago