0 ^ ∞ , It's What You Think

In this video, we explore the perplexing world of zero to the power of infinity. Is it equal to zero or does it approach infinity? To answer this question, we delve into the concept of limits and discuss how they relate to calculus, real analysis, and mathematical foundations.

We also examine the properties of limits and apply them to various examples to gain a deeper understanding of this mysterious phenomenon. Along the way, we discover that the answer to zero to the power of infinity is not as straightforward as we might think and requires a nuanced understanding of mathematical principles.

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

[BriTheMathGuy]

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[MichaelClark-uwex]

If I had an infinite amount of nothing then mutiplied it an infinite number of times, I'd still have an infinite amount of nothing.

Makes perfect sense.

[ssaamil]

"Zero to hexagon" Good one :)

[coolsongstatus]

Can u plz make vedio on proof of 1+1=2

[tarsofelix]

Interesting, so 0 ^ ∞ will always aproach 0...

I guess it makes sense, considering any x, such that 0<x<1, will become smaller each time it's multiplied by itself.

[spaghetti]

The way I figure out if an expression involving two numbers (infinity included) is indeterminate is to let one part be exact and let the other part approach its value. After evaluating, you can switch the roles. If both interpretations give the same result, it is determinate.

For example, consider 1^inf. If we replace 1 with a number approaching 1 from the right, putting it to the power of infinity would make it go to infinity. On the other hand, we can let 1 be exact and let the power approach infinity. This results in 1. These two intepretations of 1^inf lead to different results so it is indeterminate.

If you apply the same process to 0^inf, both routes lead to zero. This method works because we are evaluating the two most extreme limits that give the number we are trying to assign a value to. These limits are on opposite sides of the spectrum of possible values. So if they are the same, every limit is the same. Perhaps there is a counterexample, but I haven't seen it.

[euphoniumplayerable]

4:41 was like looking into the sun, transitioning from an almost all black screen to an almost all white screen on an 80 inch TV was hell

[dataweaver]

“infinity is not a number.” Not quite true; infinity is not a _real_ number. There are number systems like Wheel Algebras where infinity is a number.

[Amechaniaa]

0:07 that's 177 not 117 you silly

[dorismaric]

"One Hundred Seventeen"
The screen: 177

[stregamemorris]

Can you make a video solving the Zeta function with different methods

[cashley]

YouTube brought me to this video after ∞ ^ 0, where you said at the end you'll see me in the video about 1 ^ ∞. I cannot begin to explain how disappointed I am in the YouTube algorithm.

[heliocentric]

Using Brian's idea we can "define" 1^∞ to be 1, because lim(1^x) 'as x approaches infinity' is 1

[alyanshaikh]

"0^infinity is what you expect"
My mind Fuck Gotta watch this

[TopRob]

0^117 is 0 but what is 0^177 ?

00:05

[potaatobaked]

Alright, because of 3:53 now I'm going to try to rigorously define x^⬢

[realmless]

If you treat 0^(inf) as the reciprocal of zero (zeroith root) is is also determined, that is it is zero. Inf^inf is also infinity.

[teamminecraft]

I always hate that 1^(infinity) is undefined in my head it makes so much sense to be 1

[jasnoor-d-]

being a 9th grader( never done calculus or limits) didnt expect to understand it, i clicked because i was getting bored
by the way i got to say your thumbnail and titles got way better then from a year before when i remember watching you

[asc.atlantean]

if you want to strictly prove that
lim (0^x, x→∞) = 0
you have to prove that
∀ε>0 → ∃δ>0: ∀x: |x|>δ → |0^x - 0|>ε
which fails at any x<0, since 0^x is undefined there as division 0 appears


unless you specify that here's +∞, you can't say how 0^x approaches infinity: towards +∞, -∞, or even from any direction on the complex plane
therefore, strictly speaking, 0^∞ remains undefined this way