inf^0 is indeterminate

[@snbeast9545]

Are you saying "Chen Lu" on purpose now?!

[@blackpenredpen]

SNBeast Might as well : )

[@AndDiracisHisProphet]

covers the accent

[@ericherde1]

8:25 lim (x-> inf) [x ^(1/(log_5 (x)))]

[@jaredbeaufait5954]

Eric Herde yes

[@Absilicon]

12:26, well you prbly want e^(-infinity), so take x^(-1/ln(ln(x)))

[@PerMortensen]

Shit's real when you have to call in Chen Lu.

[@ssdd9911]

for inf^0 to approach 0, just use the third case but multiply the exponent by -1

[@foffif2011]

8:25 If you have: lim(x->inf) [x^(ln(a)/ln(x))] you get a as the answer.

[@yoavcarmel1245]

exactly

[@admink8662]

Except a = 0, right?

[@eddiemcgregor7537]

we can use a = 1/x so a = 0+

[@foffif2011]

Mr. K Yes right. I didn‘t think of that.

[@yrcmurthy8323]

Nice

[@mustardthe2ndkingoffastfoo570]

Find a function such that y+ y '+ integral of y + reciprocal of y = y^2. Challenge to you

[@asificare7985]

Just replace all that ln(x) by log_5(x) and e by 5 as x=5^log_5(x), then we get infinity^0-->5. Anyway, nice video and keep up the good work!

[@himanshumishra6253]

Hey blackpenredpen please upload the proof of the formula that you used in solving the problem x^2-5=√5+x... (For the -ve one)

[@gergodenes6360]

Yes, I want to see that too.

[@matejpesek2465]

Check his feed, he linked a proof from some other youtuber

[@AndDiracisHisProphet]

patience

[@himanshumishra6253]

@@matejpesek2465 hmmm... Thanks....

[@unknownvulture6189]

Yhihrndj. Eju jejn jeju uxhr

[@JuanMataCFC]

12:15 u could have just canceled the e^() and ln() since they are inverse functions, rather than taking ln(∞). nothing wrong with what u did, but i feel like this is just easier :)

[@TheTedder]

x^(1/ln x) is actually equal to e no matter what x is!

[@masheroz]

Why is everyone using log5 in their solutions? I used that in my head while watching the video, and then I come to the comments, and everyone is using it!

[@goldenshadow1775]

e^( ln(n)/ln(x) ), where n is any number you want to get (maybe without zero) after taking limit of that equation as x goes to infinity (i think that's correct...)

[@tincan357]

Infinity is so interesting, thanks for solving hard questions relating to it

[@arequina]

Please, provide link to purchasing the shirt you are wearing. I love anything math.

[@blackpenredpen]

I created this shirt myself. I will prob. have it for sale sometimes this week.

[@unknownvulture6189]

@@blackpenredpen posmss.liisu

[@Jupiter-bi5ry]

I know this isn't very mathematically rigorous, but it seems to me we could say infinity ^0=infinity/infinity (going by the normal derivation for ^0). inf/inf=j (some j). multiply by inf, we get inf=inf*j. This is of course true for any j. Thus, indeterminate.

[@Kumar-oe9jm]

Easy to get 0 just add a minus sign to the exponent in the last case to get e ^( neg inf)=0

[@DanBurgaud]

12:30 ^0 = 0 ? I know my calc teacher gave me ZERO on that exam.... so the answer is my blank test paper.

[@leif1075]

Infinity is not a number per se..so i would assume most smart people know infinity raised to a limit is fuzzy or undefined at best? Or has a limit of 1 sure.

[@factsheet4930]

10:05 And here I thought, take its square root. I'm pretty sure it would work too and no need to take ln(ln(x))? 🤔

[@VaradMahashabde]

6:25 this figuring out part be so cool

[@tperm6695]

So the lim as x->infinity x^(1/(log base n (x)))=n, isn't it??

[@arischwartzman7100]

lim x approaches inf(x^(1/log base n of x))=n

[@purim_sakamoto]

もうなんか老師の動画見てると、e^Inxって 右左折でウインカー出すくらい普通やな

[@jagmarz]

But x^(1/log,k(x)) is just a constant, isn't it? Rewrite as (e^lnx)^(lnk/lnx) = e^(lnx*lnk/lnx) = e^lnk = k. So there's no limit involved, really! That's cheating!

[@eddiemcgregor7537]

12:27 let f (x) = ln (x) / ln (1 / x) lim (x-> inf) [lim x ^ (1 / f(x))] = 0+

[@plislegalineu3005]

12:17 I'd just cancel e and ln but thats not wrong

[@mathsdiscoveries9221]

You can have many ways to do it and still get a number. Why? Because infinity equals 1/0, and if infinity is powered to 0 it will make the fraction (1^0)/(0^0). As you can see, 0^0 actually equals to every number. If 1/(every number), you will always get a number.

[@zeldasama]

If lim as x approaches infinity for x^(1/lnx) goes to one is it safe to assume this. The limit as x goes Ti infinity of x raised to 1 over f(x) will give you the answer as such. Whatever makes funtion f(x) 1 is the answer so. So therefore, the limit as x goes to infinity of x^(1/f(x)) where f(x) is x²+2. Lim x>inf x^(1/x²+2) is i, as in √-1. I get this answer by putting the denominator of the power equal to 1 x²+2=1 which is same as x²=-1 so then x is i. Although, this would give plus or minus i, so I am feeling I made a false assumption.

[@jonasharestad7664]

The last example you gave times 5/e :)

[@chixenlegjo]

Now make it equal a negative number.

[@VolodymyrLisivka]

Logical error: limit of something, approaching 0 at infinity, doesn't equal to zero when multiplied by infinity. So, inf^0 == 1, but inf^(lim(x->inf)[something] -> 0) can be anything, because infinitely small value still has value. It's like x/inf - it's zero for all x except x=inf, but we cannot say that inf*0 !=0 because (logical error) x/inf == 0. So inf*0=0, inf^0=1, unless you substitute 0 with something else.

[@shawnpheneghan]

Click Bait! Infinity is a limit concept. 0 is not. In this video infinity is not raised to the power 0. I believe that is equal to 1.

[@thexoxob9448]

For limits, do know for the fact that the value would never equal the number that it approches

[@officialredslimecraft4092]

Just change the base on the log to anything :D so that when the x in the base converts to k^logk(x) it ends up as k^1

[@ChrisMMaster0]

So the general formula would be LIM x-> inf [ x^(1 / log_n_(x)) ] where n = any positive number; and you would change the base x to n^(log_n_(x)) to get the desired n value.

[@jacobpinson2834]

I think this is more a proof that 1/infinity is not equal to zero, but a number infinitely close to zero.

[@unbekannthi7935]

X Go the log base 5 of x power

[@ВасилёкСаныч]

mắc cười quá haha

[@plislegalineu3005]

if you want inf^0 = a , when a is finite and positive, just do lim x->inf x^(1/log_a(x))

[@theophonchana5025]

infinity^(0) = undefined

[@markobrenchuk9947]

When x->0 Lim(1/(x^(1/x))) = e^(-infinity) = 0 That how Infinity^0->0

[@Hari_Om12]

So I guess the video suggests that there is no general solution to expression infinity raised to the power 0.

[@brunolevilevi5054]

infinity^0, 0 times infinity, 0/0, infinity/infinity and 1^infinity are all indeterminate. If you encounter a limit like that you can't draw any conclusions. If you find something like 5/0, the limit is either going to be positive infinity or negative infinity, depending on the way you approach 0 (from the right or from the left). If you find infinity + infinity you can also deduce its going to be infinity without having to use any calculus or anything like that

[@ИванКальянов-л4с]

Bruno levi Levi you can still find the limit using some other stuff as he did in the video

[@Cloud88Skywalker]

@@brunolevilevi5054 to complete your explanation: This only applies when those 1s and 0s appear as a result of a variable approaching something. Otherwise there's no indetermination.

[@samhughes2957]

The answer to the final challenge, I believe, centers around having the exponent approach negative infinity in the same way he had the exponent approach positive infinity in the last example. I may be wrong but I think x^(-1/ln(ln(x)) should do the trick. After you change x to e^ln(x), multiply the two exponents and take the limit and derivatives of each, you should have negative ln(x) as x approaches infinity. Thus e^-infinity =0

[@wil-fri]

Why am I watching to an asian guy talking in english about maths?

[@MathIguess]

Make the first example 5(x^1/x) to end up with 5 :P

[@rshawty]

8:07 but yi is one so ...

[@leoyang1.618]

Hey blackpenren please prove the answer to zeta(4) or zeta of a positive even integer.

[@lordofcastamere9376]

Instead of 1/ln(x) i think 1/Log5(x) should give the answer 5

[@tonyhaddad1394]

Infinity^0=infinity/infinity so is not equal 1 indifine

[@srpenguinbr]

It is undetermined because inf^0= (e^ln(inf))^0 e^0•inf

[@126sivgucsivanshgupta2]

What do u mean by smaller infinity ??

[@Ben-wv7ht]

could (x^x)^1/x also work to proove it can be equal to +inf ?

[@gergodenes6360]

Infinity to the 0th power approaching negative infinity? :'D

[@admink8662]

lmao

[@noonesherem8782]

In FUN ity

[@michaelsidorov5508]

Sophistic.

[@EnPee91]

To get any number y>1, you can use x^(1/log_y(x)). And to get any number 0

[@albertoaf5301]

😅😅

[@albertoaf5301]

😅😅

[@ΜαρίαΧριστονίκου]

sometimes it is undedectable.It cannot be evaluated

[@unknownvulture6189]

Hi Brooke

[@MrHK1636]

You get 5 from limit lim x->infinity x^(ln x/ln 5) You can always do a^log a(x) =x

[@peanut12345]

e is not equal to 1, euler said is 2.7.

[@iironieminen6849]

lim x->infinity x^(ln(a)/ln(x)) = a

[@rimanbachar7588]

Plz Explain 0^0=?🤔🤔🤔

[@AndreaDAlu]

inf to 0 power is like lim x to inf of x/x l’hopitals rule 1/1

[@AndreaDAlu]

btw i know it isn’t the correct answer but

[@UPSE-qd2ql]

Where can I buy your t shirts ? Love them

[@anirudhnarasimhan7307]

Nice video

[@diabolicallink]

11:50 niice

[@jaysn1683]

3:42 What is that rule called?

[@sabinrawr]

L'Hopital's Rule.

[@user-wf2fm1yj4k]

You need some very basic calculus to use it

[@DL-uo3xt]

How beautiful and interesting!

[@Rtong98]

Hahaha genius! Awesome video

[@LaurentBessondelyon]

Don't find and don't know PETA's rule.... Any help?

[@rubenkingcard3874]

Laurent Besson it is l hôpital Rule !

[@LaurentBessondelyon]

@@rubenkingcard3874 but yes it comes back to me ....😋

[@LaurentBessondelyon]

@@rubenkingcard3874 and probably I did a mistake on the word petas.... Perhaps he said hosPITAL law.... 😢 My fucking English.... 😋

[@Axem26]

for the ∞^0 = ∞ what about x ^ {2/[ln(x)]} ?

[@encounteringjack5699]

Axem26 that wouldn’t work. [e^(ln(x))]^(2/ln(x)) = e^[2(ln(x))/ln(x)] = e^2 Which is just a constant so the limit will just be e^2, at least by that method. Which is why I say it wouldn’t work.

[@ДмитрийФедоровичев]

now i have nothing to believe in

[@mandaglodon]

Proof X+1/X =2

[@Quwertyn007]

Umm... 5^x^(1/x)?

[@afafsalem739]

Great video

[@toripuru0069]

256 VIEWS IM HERE

[@DjVortex-w]

I want inf^0 = i

[@flamingpaper7751]

This isn't necessarily related but how would I solve for Z if Z^Z=-2?

[@Apollorion]

Think complex & start by taking natural logarithm on both sides.

[@flamingpaper7751]

@@Apollorion Zln(Z)=ln(-2). Now what?

[@Apollorion]

Let's write Z as z=r*e^(i*t) and ln(-2)=ln(2)+i*pi*(1+n*2), at which n is any integer and r and t are the polar coordinates of z (that is: think complex). Then we'll get: r*(e^(it))*(ln(r)+it)=ln(2)+i*pi+i*2pi*n Know the formula of Euler (e^ix=cos(x)+i*sin(x)) and make the real and imaginary part of that complex equation equal at the same time. Good luck.

[@Apollorion]

@@flamingpaper7751 I've puzzled a little further & I think I've 'solved' it; prescriptum: a) Z^Z is used -> I'll probably need to use the Lambert W function. b) ln(-2) isn't defined as a real number -> I'll need to think in complex numbers. c) According to en.wikipedia.org/wiki/Lambert_W_function the Lambert W function is also defined for complex numbers. -> A solution might be found. -> Let's give it a try. Z^Z = -2 ln(Z^Z) = Z*ln(Z) = ln(Z)*e^(ln(Z)) = ln(-2) = ln(2)+i*pi+i*2*pi*n W(ln(Z)*e^ln(Z)) = W(ln(-2)) = ln(Z) = W(ln(2)+i*pi+i*2*pi*n) Z = e^W(ln(-2)) = (ln(2)+i*pi*(1+2*n))/W(ln(2)+i*pi*(1+2*n)) *n* can be any integer, i.e. infinitely many solutions, because for each solution of *e^x=-2,* is *e^W(x)* a solution to Flamingpaper's equation *Z^Z=-2.*

[@neilgerace355]

Good old Chen Lu

[@DaanSnqn]

USE THE CHEN-LU!

[@skilz8098]

According to my TI 84 Plus calculator when I put in the expression: (E99^0) where E99 is representative of +infinity it, sure enough, gives me the value of 1! It was always to my understanding that any number raised to the 0 power except for 0 is 1. Considering that the graph of the equation y = x^0 is approximately equivalent to the graph of y = 1 except for when x = 0, I think I'd still stick with the fact that infinity^0 = 1.

[@TazPessle]

Any (x^1)/(x^1) = x^0 If you're using equivalent infinities then the answer has to be 1.

[@KineticManiac]

The main problem with your logic is that, when drawing the graph of x^0 and basing your conclusion on that, you're essentially considering the limit of x^0 as x approaches infinity. Which is 1. But that's only one of the possible ways of thinking about infinity^0. Drawing a different graph results in a different answer. In fact, this is the same reason why 0^0 is also considered indeterminate. Depending on how you approach it (no pun intended), you will get different results.

[@asi3136]

nice

[@sagnikadak1498]

1st comment

[@andi_tafel]

(1/x)^x approaches 0 if x approaches 0

[@gergodenes6360]

That turns out to be lim 1/(x^x) as x approaches 0, which is 1/(lim x^x as x approaches 0), which is 1/1 = 1, you messed up somewhere calculating that.

[@andi_tafel]

Hm, you're right. I did it using ln but in the end I forgot to do e^0 so I got to 0 instead of 1.

[@exo_01]

@@gergodenes6360 Actually the limit of x^x as x goes to 0 does not exist, because the left-hand limit, namely the limit of x^x as x goes to 0 from the negative side, which equals - 1, does not equal the right-hand limit, which, you named it, is indeed 1. So talking about "the limit" of x^x as x goes to 0 does not make a lot of sense, it's just the left- or right-hand limit that exists. Cheers.

[@gergodenes6360]

You can't even calculate it from the left hand side if we wanna keep it in the Real world

[@gergodenes6360]

And actually: www.wolframalpha.com/input/?i=lim+x%5Ex+as+x-%3E0

[@aGuyWithConscience]

You have made a clear thing confusing. As to the 3 powers, x>ln(x)>ln(ln(x)). When x's and their powers approach to infinity simultaneously, x goes much faster the ln(x) and ln(ln(x)). That is the infinity of x is much larger the those of ln(x) and ln(ln(x)).

[@nathanisbored]

the second one feels like cheating because its not really infinity^0. the expression is literally e so theres no reason to plug in the limit. its like saying: 1 = x^0 plug in infinity you get inf^0 that shouldnt really count

[@danielbenyair300]

X^0=1 always!!! unless there is method error

[@danielbenyair300]

1:20 wrong... infinitesimals are NOT zero!!! It is the infinite root of infinity 5:53 this means you CHOOSE the answer yourself!!! therefore it is wrong!!! 6:55 6times intinite is BIGGER then 5 times infinite!!!! 8:50 i stoped here...

[@blackpenredpen]

Daniel Ben Yair : )

[@danielbenyair300]

@@blackpenredpen i would like you to do a video about 3n+1 and fucos on the numbers where the lines connact (start with powers of 2)