How the fuck does this work? Explain

How the fuck does this work? Explain.

I refuse

It wants to say that: if you have a sequence a_n which goes to 1 and a sequence b_n which goes to infinity, then you can't say anything about a_n^b_n with that information alone.

If a_n= 1+1/n and b_n=n you get e.
If a_n=1-1/n and b_n=n you get 1/e.
If a_n= 1+1/n^2 and b_n=n you get 1. (same with -)
If a_n=1+1/n and b_n=n^2 you get infinity.
If a_n=1-1/n and b_n=n^2 you get 0.
If a_n=1 and b_n=anything you get 1.

Hint: [math]a^b \,=\, \mathrm e^{b\,\ln\,a}[/math]

1^inf = e^(inf * ln(1))
1^inf = e^0
1^inf = 1

Now suppose 1 is a limit.

>inf * 0 = 0
"""""""no""""""

As 1 approaches 1, 1 is 1?

So if you add enough 0s eventually they stop being 0?

just read your lesson and try to understand that you're doing operations on limits, not numbers

As [math]x[/math] approaches 1, [math]\ln\,x[/math] is 0.

>eventually stop
that is the thing, it doesn't stop.

Hah, that's a nice, short way to put it.

>As x approaches 1, ln(x) is 0.
So wait, how did you get to having an x and comparing it to 1 in the first place? I feel like I missed some steps. We're starting from here:
1^inf = e^(inf * ln(1))
Do you rewrite the equation in some way to get there? Like:
x = e^(inf * ln(1))
ln(x) = inf * ln(1)
ln(x) = inf * 0
Where is the approaching 1 part happening?

indetermined forms are stupid and lame.
The real meaning of this is that if you add an extra elements to the set of reals and call it "+infinity", and put in it the obvious topology (*), then there is NO WAY to build a continuous map F from [0,+infinity]^2 to [0,+infinity] which is both
(i)continuous (**)
(ii) for every a,b real (finite) numbers we have a^b = F(a,b)

...
(*) and (**) if you don't know what these words means it doesn't matter, continuity would yield the following property: for every sequences of real positives numbers (n-> a_n) and (n->b_n), if a_n has limit l (real or infinite) and b_n has limit m (real or infinite) when n grows, then a_n^b_n has limit F(l,m).

>
(*) and (**) if you don't know what these words means it doesn't matter, continuity would yield the following property: for every sequences of real positives numbers (n-> a_n) and (n->b_n), if a_n has limit l (real or infinite) and b_n has limit m (real or infinite) when n grows, then a_n^b_n has limit F(l,m).

Let us call (iii) this claim above.
>
>(ii) for every a,b real (finite) numbers we have >a^b = F(a,b)

It is *impossible* o have simultaneousy (ii) and (iii)

1^n with n being any real number would get you 1. Since infinity is not a number, this does not apply.

That's wrong.

It should be that as a number that infinitely gets closer to 1 is multiplied by itself an infinite amount of times, the answer is indeterminate

[math]\lim_{x\to1} x^\infty = indeterminate[/math]

That should be at least a little intuitive, because although it would seem like the answer should be 1, the infinitely small distance to reach 1 is amplified an infinite amount of times, so really its unknown (without some calculus) if it actually will approach 1, infinity, or some number in between.

However, if instead you have a value that's precisely and concretely 1, not a value that approaches 1 with an infinitely small distance between, then no matter how many times you multiply 1 by itself, it'll still be one. So,
[math]1^\infty = 1[/math]
and the OP image is false.

Why are you taking the limit of 1 instead of infinity???

lim x->inf 1^x converges uniformly to 1 doesn't it?

I wasn't taking the limit of 1 as it approaches infinity; I was taking the limit of a number that approaches 1 as its raised to the infinite power.

And yes
[eqn]\lim_{x\to\infty} 1^x = 1[/eqn]
is true

Obviously, but we're given 1^infinity and you're making 1 the variable instead of infinity and I'm asking you why you would do that when it isn't convergent and arithmetic operations with infinity as a number isn't defined, and making infinity the variable gives us a convergence?

we're given the word "indetermined" which implies we are talking about convergence rules for function composition, so there's gonna be two occurrences of the variable
are you being dense on purpose?

You're fucking retarded, I apologize for trying to assist you.

Infinity is usually implied as a limit, while 1 typically isn't thought of as that. I was just trying to explain for the OP the easiest way to get confused in that sort of problem.
If you want to get semantic, arithmetic operations with infinity as a number aren't defined because infinity isn't a number. However, limits to infinity are well defined.

infinity is not a number
/thread

Putting a bunch of symbols together doesn't give them meaning. 1^infinity clearly doesn't mean anything, since a^b is a*a*a... for b times total. How could you determine a^infinity given that definition? You can't, because infinity isn't a number.

Does putting random words together give the resultant sentence any meaning? Not necessarily. Stop being a brainlet

you could raise 1 to the burgerth power and it'd still be 1

Why do retards always get so hyped up by things like this? Are you on acid? Do you think there's some penetrating insight into these meme definitional things that only you can see?

Fuck off.

There will come a time in your math career when you have to use limits. Limits allow you to become arbitrarily close to a number WITHOUT it actually being that number.
This allows to you show how 1/0 is impossible because it will approach infinity as you get extremely close from the positive side and it will approach negative infinity from the negative side. A limit only exists if both of the limits from each side of the value approach a finite number.
When dealing with infinity, you almost need to use limits in the process to create viable math because you cannot be infinity you can only approach it.

>the burgerth power
0.56?

>using infinity like a mathematical constant and attempting to do arithmatic with it