x^sqrt(x) when x is 0 vs. x is approaching 0

[@jzanimates2352]

Black penned pen, I got moved up to the top class in maths, all thanks to your videos! You are an awesome KZbinr!

[@blackpenredpen]

That's so awesome!! Thanks for letting me know!! Great job to you as well!!!

[@bruhmomenthdr7575]

​@sunnygames4003Black penned pen 🧐🥸

[@rosebuster]

I think that to be perfectly formal when using method 1, we'd first have to prove that the limit exists in the first place before we can call it L. Otherwise we're doing calculations using a value that's not a real number. In this case it didn't really matter, but I think things could go wrong in other situations.

[@itsiwhatitsi]

You and 3Brown1Blue are the best KZbinrs for Math. You teach things better than in school. Very good job, greetings from Italy.Peace

[@NikoThePancake]

Bprp vs 3b1b Math-off

[@kaiandchanellesimmons502]

Well done for 200000 subs. I just realised. I'm a uk sub currently doing maths at what we call A-Levels. You have helped me so much and want to say a huge thank you.

[@blackpenredpen]

Kai and Chanelle Simmons you're very welcome, thank you for your nice comment!!

[@SartajKhan-jg3nz]

'In your exam do not write "DO MORE WORK", do the work and give the answer' 😂😂

[@inyobill]

Brings to mind that classic cartoon: "... and then a miracle happens ...".

[@nootics]

Love your videos, I'm just at a "high school" and watch them for fun. My mathematical understanding got so much better as well, even though I haven't had most of what you do in your videos in class

[@knoobiez]

Happy college

[@knoobiez]

0:30

[@Fire_Axus]

your feelings are irrational

[@gaurangagrawal6251]

Me : No the answer can't be zero it must be 1. Bprp: calm down dude we have found lnx to be zero .

[@waishingtseung6930]

What?

[@Kumar-oe9jm]

At 8:46, are u sure that u can multiply both the top and botton by 2x^(3/2) when the limit is x approaching 0+, bec it is like saying 2/3 multiply by 0 on the top and bottom and saying it equals 0

[@angelmendez-rivera351]

Kumar No, because it is a limit, so you are never actually multiplying by 0

[@Vibranium375]

You haven't invoked the limit yet

[@Inspirator_AG112]

0⁰ = 1 doesn't only work as a limit, but as an actual value.

[@superhil4487]

nah, 0^0 = 1 is by a definition, it's similar with 0!=1

[@angelmendez-rivera351]

Well, y'all know already what I have to say about the "0^0 is undefined" claims. The mathematics pretty unambiguously show, under any reasonable definition of exponentiation, that 0^0 = 1 follows directly from that definition. THAT BEING SAID, I 100% agree with BPRP in that you should avoid confusing 0^0 (where 0 here is exactly the natural number 0, or the real number 0, whatever you prefer) with (0+)^(0+) (where 0+ is actually just really bad notation here for lim δ (δ -> 0+)). The former is an arithmetic expression, with 0^0 = 1, while the latter is a limit expression, and constitutes an indeterminate form. This means that lim f(t)^g(t) (t -> c), given the restriction lim f(t) (t -> c) = lim g(t) (t -> c) = 0, can be equal to 0, or any positive real number, or even perhaps +♾, depending on the exact relationship between f and g. Notice that this limit is not the same thing as 0^0, which again, is simply an arithmetic expression, where, as BPRP says, "the 0s are exact 0s, not limits."

[@derwolf7810]

Offtopic here, but as i had issues in correcting my post (had to recreate that post). I wanted to thank you for pointing out that error (and let you know).

[@nishadpatkar6636]

At 6:20, wouldn't it be easier to raise both sides to e so u get e^sqrt(x)+x and since x approaches 0+, you get e^0 + 0 which is 1?

[@Zelda1990s]

While it is true that 0^0 is an indeterminate form, I thought out of "convention", we say 0^0 is 1.

[@angelmendez-rivera351]

Calling "0^0" an "indeterminate form" is to fail to understand what an indeterminate form actually is. Here is the simplest version of the argument: lim x^y (x -> 0, y -> 0) does not exist. Therefore, the value of lim f(t)^g(t) (t -> c) cannot be determined from the fact that lim f(t) (t -> c) = lim g(t) (t -> c) = 0 alone: you also need to have information relating f and g specifically. This inability to determine lim f(t)^g(t) (t -> c) from lim f(t) (t -> c) = lim g(t) (t -> c) = 0 alone means that this limit, with these restrictions on f and g, constitute an indeterminate form. Note: this is NOT the same as 0^0. 0^0 is not a limit. 0^0 is a numerical expression, and the value of this numerical expression is 1, which can be proven from the definition of x^y directly. Limits and forms have absolutely nothing to do with it.

[@sabinrawr]

Why can't we have a complex exponent? Why doesn't it make sense to have a negative base to a function power? As for solving problems involving the real world, you are right of course. But couldn't it still be done mathematically? Couldn't it at least be a little bit interesting to see what would happen?

[@mike4ty4]

You can, indeed, have a complex exponent, and raise negative bases to real powers. The most common example is the imaginary exponential e^(ix), which is frequently used as a description of waves. The trick is that if one uses negative/complex bases that the exponentials in general become ambiguous thanks to the ambiguity of the complex logarithm, and thus one has to specify what "branch" of the logarithm one is using for a given problem. Because of this, it is customary to stick in most cases to the typically more well-defined e^x (or e^z, for complex) with suitably-structured exponent.

[@tipoima]

After plugging -0.000001 in Wolfram Alpha, it seems like it converges on -1 (the answer was something like -0.9....+0.00001i)

[@Rudxain]

@@tipoima That's strange, when I put the actual limit approaching from the negative side, I got the same answer as from the positive side, +1 not -1

[@JindraAG]

I think an honest answer is you can, but at that point we've moved past simple calculus.

[@mybarca8083]

You can also make an educated guess as if x 0.25 and x^y being 0.5 in this case, if you fill in a smaller number like x=1/25 and fill in 1/5 for y then this will give you 0.53 thereby it can be assumed that when x approaches 0 and y approaches 0 as well, x^y will give you 1

[@Sgrunterundt]

Here's a challenge with a satisfying solution that I just figured out myself: The function f(x)=x^x^a has one global extremum for a != 0 (a minimum when a>0 and a maximum when a

[@jacobpinson2834]

when the blue pen comes out, you know real maths is going down

[@Cloud88Skywalker]

My intuition would have been to let both parts of the indeterminate form compete and see which one wins: 0^x = 0 x^0 = 1 √x will get to 0 much faster than just x, so f(x) inclines to look like x^0 rather than 0^x, so the answer is 1.

[@Cobalt_Spirit]

You say that as if the indeminate form 0^0 could only end up as either 0 or 1. Not true.

[@Arthur-so2cd]

@@Cobalt_Spirit root x is getting smaller much faster than x, therefore you have a small number to an even smaller number

[@Rudxain]

I we want to get fancy, we could allow Dual and Complex numbers. The 2nd solution to sqrt(0) is called the "epsilon unit", so directly plugin it we get 0^e (the "e" in this context is not Euler, it's epsilon). For the limit from the negative side, we get the same answer as the positive side on WolframAlpha

[@davidseed2939]

Note the google calculator gives 0^0=1 so is there a function which is of the form 0^0 at x=0 but is not=1 as x->0+

[@sunsetflory]

i thought about the 2nd way to solve it! we usually work in class with that form

[@orisphera]

Fun fact: 1 is also the most logical way to define 0**0. That's what we get when we approach it in a straight line other than the line of constant base and that's also what it is in combinatorics. The original definition gives that because the product of 0 numbers is 1 and in my version of matrix theory (I'm pretty sure someone else came up with the same one) that's what the determinant of a 0x0 matrix 0I (which is the only 0x0 matrix) is

[@MuffinsAPlenty]

Yes, the determinant of the 0x0 matrix is, indeed, 1. And this is _completely_ consistent with all of the theory of linear algebra and abstract algebra.

[@yee3816547290]

Function value is different from limit. The two need not to be equal. 0^0=1 is the only reasonable definition.

[@gilber78]

BPRP: I have a suggestion for “so you think you can take the derivative” Can you do d/dx of log base x of a (where the input of the log is a constant and the base of the log is a variable)? I know the answer- I’m curious how you’d do it in a video

[@NoNameAtAll2]

log(a,b)=ln(b)/ln(a) problem?

[@archithtelukunta4599]

Log base x of a can be written as loga/logx(its an identity).Therefore the derivative is same as the [derivative of 1/logx]*loga

[@gilber78]

@@NoNameAtAll2 I know but I just wanted bprp to do it because I know the answer already. I know the identity

[@yee3816547290]

0^0 is different from limit of 0^0 form. Don't be confused.

[@DanBurgaud]

How I wished YT was available 40 years ago... LOL!

[@Lasersharp]

Is it true that 0^0 is undefined? I thought that was the case for a while, but some people say it's actually defined to be 1 (with valid arguments from a combinatorics perspective, as 0^0 is similar to 0!).

[@angelmendez-rivera351]

Lasersharp For the purposes of undergraduate mathematics, 0^0 is undefined just as the logarithm of a negative number is undefined for students in algebra 1.

[@angelmendez-rivera351]

Okay, I am revisiting this 2 years later, and I have no idea why I said the nonsense that I said. 0^0 is unequivocally not undefined. 0^0 = 1 even for the purposes of algebra 1. It follows directly from the definition of exponentiation.

[@atmonatmon2947]

Shoot a video about what is t: a ^ b = b ^ a * t

[@JayTemple]

I used the exponential form, but I replaced x with 1/u and let u go to infinity. I also didn't have to invoke "ln L equals," but I'm not sure that the former enabled the latter.

[@OculusOfficial]

Depending on which pattern you want to infer its meaning from, 0 to the 0 will be whatever you want, I prefer to think that multiplying 0 by 0, 0 times we will get the result 0. The reason we say undefined I think is to satisfy the people who want to leap from a pattern to a solution without proof by not saying that they are wrong but by saying that we dont know.

[@angelmendez-rivera351]

This is just wrong. 0^0 is not undefined, and most mathematicians acknowledge that 0^0 = 1 when they write their proofs, and even some textbooks acknowledge it as well. There are only a handful of mathematicians who refuse to accept the fact that 0^0 = 1, but this is just a petty denial of reality, no different than when high school students deny the fact that 0.(9) = 1 even after seeing rigorous proofs. As for BPRP or other teachers saying 0^0 is undefined, I have no idea why they do so. I have had conversations about this with BPRP in the past, and the arguments almost always boil down to the fact that lim x^y (x -> 0, y -> 0) does not exist, which is not actually a valid argument, because the definition of a function at a point is not contingent on the limit of the function to that point. If you simply substitute x = 0, y = 0 into the definition of x^y, you get a simple, unambiguous answer, and the fact that arguments from limits are being used to pretend that this is not the case baffles me and irritates me. Honestly, you may be right about one thing: it does feel as though the only reason teachers say 0^0 is undefined is solely for the purpose of making the minority of mathematicians who agree with them happy. However, I am not so dishonest as to claim to be certain that this is the reason behind everything, or that there is a reason at all, and I could just be misunderstanding what is happening. Regardless, I think this type of nonsense, where teachers teach things that are directly in contradiction with what mathematicians study and know is unacceptable under any circumstances. It is one thing to oversimplify a topic and introduce it infornally to students rather than rigorously so that they can get an intuition for the topic, it is a different thing entirely to outright lie to students.

[@MrDazzlerdarren]

Is 0+ kinda like 0 + dx? If not how does it differ?

[@dudurododoizi8547]

can we have a 0^0 situation where it's not equal to 1?

[@eric3813]

Yes

[@zblxst9347]

Yes we can : lim(x-->0) 0^x = 0

[@IoT_]

@@zblxst9347 only right side limit or zero plus. You can't evaluate limit as x goes to zero minus of 0^x

[@dudurododoizi8547]

@@kyro1197 oh thanks I didn't see till end

[@zblxst9347]

@@IoT_ Obviously we can't. I didn't think about it but it's a good precision

[@GOLDman4856]

0^0 is actually equal to one as a computational question. One way to prove this is using blackpenredpen's "BEST FRIEND" formula

[@angelmendez-rivera351]

Exactly. Using limits cannot disprove this.

[@johnbohnenstiel605]

could you please do the sum of the reciprocals of Mersenne numbers (3,7,15,31,63,...). It converges to about .606. Is the closed form 1/(sqr root of e)? Something else? Please show me! Thanks- love your content!

[@nanamacapagal8342]

Wolfram showed up with something about (log(2) - q-digamma(1/2, 0, 2))/log(2) anything but a simple 1/sqrt(e)

[@tarekhajjshehadi4670]

What is (i!) ?

[@jjeherrera]

Do you mean (i)! ? If you do, my guess is it would be an alternative notation for the gamma function of 1+i, in which the expression for an integer is used for a complex number. (Gamma(n+1)=n!). Just a guess...

[@angelmendez-rivera351]

jjeherrera Black Pen Red Pen already addressed this

[@JeanSarfati]

Thanks. I begin to get the symmetry between ln and e ! It's basic but well, now i visualize the two symmetric curves... Visualization in on a snap, developing algebra is like the speech, step by step then rigorous.

[@otakurocklee]

Hmmm... the first method doesn't work because ln x is not continuous at x=0. So you can only bring the limit inside if you know that lim x->0 x^(sqrt(x)) is non-zero. We don't know that it is beforehand. Second method works before e^x is continuous everywhere.

[@maxwellmogadam399]

ive always thought about sqrt(0) this is a great video thanks!

[@rafciopranks3570]

I have an equations X^X+X=1

[@ssissi12]

hum 0 ?

[@rafciopranks3570]

Yes but... there is one more answer :)

[@Fahrradnerd]

0? 0^0 is undefined so I would argue 0 is no solution

[@ssissi12]

@@rafciopranks3570 yeah but i think it's an irrational number x=0.3036591268776...

[@angelmendez-rivera351]

Marcel No, 0^0 is generally taken by mathematicians to be 1.

[@Christian_Martel]

Could you calculate the same limit, but from the left hand side, lim f(x) where x -> 0- ? I would “imagine” this is possible.

[@hugo3222]

If you did x^x before, just do x^sqrt(x)=(sqrt(x)^sqrt(x))^2=(y^y)^2 (with y=sqrt(x)), and you're done.

[@Peter_1986]

I once spent like 30 minutes trying to solve a very weird example problem, and then the solution said that a solution didn't exist.

[@madanmohan877]

Please clarify: If y = g (x)/h(x) then y= g'(x)/h'(x) . Is it correct. That is to say: differentiating numerator and denominator simultaneously does not spoil the equation like multiplying the numerator and denominator with the same number. Please clarify it.

[@angelmendez-rivera351]

Search "L'Hôpital's Rule"

[@नवलशर्मा-ड7ज]

Black pen red pen please make a video on x^1/lnx as x~0.👍

[@frankharr9466]

Yay! You know, when I was taking these classes and they told me to go and do work when I could intuit the answer, I didn't realize that they were trying to teach me techniques. Can you do approaching 0 from the i? Nah, that would be silly.

[@Mathelite-ii4hd]

what happened at 8:38?you said it will be -1/2x^-3/2 but then you wrote it down -1/2x^3/2

[@adamp9553]

I could answer this question pretty immediately, without any calculus, understanding both (0+)^(0+) and how square roots work with real numbers. (0+)^(0+) is already 1; and a square root brings a term closer to 1; despite the power becoming closer to 1, lim with an infinitesimal will still approach x/x, which makes 1 for any non-abstract number.

[@AngadSingh-bv7vn]

We can do the same thing for x^x and we'll get the same thing so lim as x->0+ of x^x is also one.

[@hellohabibi1]

In WolframAlpha when you enter Limit[x^sqrt(x),x->0] it gives 1? Why is that? How does 0 negative also go to 1?

[@mike4ty4]

One of the things I'd want to point out is there's actually nothing weird or mysterious about "indeterminate forms". Rather, they are simply singularities of the arithmetical functions. For example, 0/0 is a singularity of the two-variable function ("binary function", which is why that / is called a "binary operation") f(x, y) = x/y at (0, 0), and it is a non-removable singularity, so there is no way to extend it continuously to that point. In other words, the two-dimensional limit does not exist. A one-dimensional limit giving the 0/0 form represents a certain path of approach to that singularity, and thus its value depends on the specific path taken, meaning you have to treat that limit individually, not generally. The same goes for 0^0, though there seem to be a bit of consensus that it should be defined to be 1. Nonetheless, that doesn't change its status as an indeterminate form since f(x, y) = x^y is still _not continuous at (0, 0)_ , and thus you are proscribed the use of direct substitution, just as for any more "obvious" discontinuous function that hasn't been graced with this mysterious "indeterminate" label. Unfortunately there are a lot of very helpful notions like this that get glossed over in much maths teaching dogma as embodied in the textbooks, and I think it's worse off for it. Rather one is just told that certain forms can't be used, without much inkling as to _why_ or what the notion of "indeterminate form" even _means_ , and, moreover, this shows there are many other forms than just the ones listed: _any_ non-removable singularity of a polydimensional function (or n-ary, if you prefer) is an indeterminate form when a lower-dimensional limit of some composition of that function yields it.

[@cycklist]

LOL some people just love to show off, don't they.

[@mike4ty4]

@@cycklist I thought this was a mathematics site, where we discuss about maths and thus that a detailed post about a piece of maths related to the specific subject matter of the video would be welcomed. The point was to try and provide information some might find useful, given that this is another one of those things (kind of like the fact that an indefinite integral with disconnected domain, like the integration of 1/x to "ln |x| + C", actually effectively has more than one constant of integration) that isn't necessarily mentioned even though it should be and would remove some apparent arbitrariness in the exposition of the material. By posting this, it may help some to understand things they didn't before due to possible shortcomings of the usual presentations. Not to "show off". Don't just assume you know what motivates people, especially those you don't know very well. You could be wrong. And while this is a rather trivial matter in the grand scheme of things, this same _thinking_ and approach to people can cause real hurt in the world in other circumstances. It can be the origin of prejudices (like racism), of other-wallet worrying, and ultimately perhaps, even of wars. Keep that in mind, and learn to be more Good, since that's what the real purpose of human life is, at least I believe, from a metaphysical point of view.

[@MuffinsAPlenty]

mike4ty4 - This is probably the best comment I've seen on this video. It's a shame that so many people are willing to throw their understanding of limits in the trash when it comes to talking about defining 0^0.

[@angelmendez-rivera351]

I am 2 years late, but this is a big agree from me. In fact, this comment was very illuminating to me, and one of a few speeches that sent me down on a quest to re-educate myself in mathematics and free myself from the problematic mathematics-education dogma that plagues people on a seemingly world-wide scale. This comment should be pinned.

[@pedrosso0]

I am proud of myself, I did everything here except setting the limit as L and just putting it as e^(limit)

[@pedrosso0]

Trying to do the limit as x goes to 0 minus. It's hard because of the imaginary part but I think the answer is that it goes to 1

[@nanamacapagal8342]

3rd method: do it numerically. Compute 0.0000000000000001^0.00000000000000005 4th method: graph it.

[@neoss988]

0^0 is undefined? ok Let's define it : I put 0^0 = 1 X)

[@vaniragujana]

Nice explanation. It is true that both X^x and X^sqrt(x) functions are equal to 1 when lim x->0; (i.e.x->0+) Please clarify!

[@erikosterling3311]

yes

[@terapode]

And as always, a great video.

[@markgraham2312]

Try considering f(x) = |x|^x or f(x) = |x|^sqrt(|x|), then you can take lim x-?0- .

[@tonyotag]

if the limit was coming from the left hand side would it not be a version of sqrt(-1)? or a version of imaginary numbers? so limit x^sqrt(x from 0- side) then is imaginary number answer?

[@angelmendez-rivera351]

tonyotag Not quite. The exponent will be imaginary, but the overall power is complex-valued. You can evaluate it using polar coordinates more easily.

[@tonyotag]

@@angelmendez-rivera351 I would "imagine" so. My comment was based more on blackpenredpen's comment in video about sqrt(x) is negative and therefore cannot exist, it does exist per immignary number i = sqrt(-1) in the exponent

[@angelmendez-rivera351]

tonyotag The limit was presented as a problem of real-valued functions, so imaginary numbers are not allowed.

[@JoshuaHillerup]

Is there a situation where you have (0+)^(0+) not equalling 1?

[@angelmendez-rivera351]

Careful. 0^0 = 1, but 0+ is just bad notation for expressing lim ε (ε -> 0), so (0+)^(0+) is bad notation for expressing lim x^y (x > 0, y > 0, x -> 0, y -> 0). This limit does not exist, but if you consider x = f(t), y = g(t), such that lim f(t) (t -> c) = lim g(t) (t -> c) = 0, then lim f(t)^g(t) (t -> c) can be equal to any positive real number or to 0, depending on the specific relationship of given f and g.

[@MichaelRothwell1]

An example where the limit is 0 is f(x)=e^(-1/x²), g(x)=x, so f(x) & g(x) -> 0 from above as x->0+, and f(x)^g(x)=e^(-1/x)->0 as x->0+.

[@MichaelRothwell1]

An example where the limit is a (for 00+, and f(x)^g(x)=a->a as x->0+.

[@MichaelRothwell1]

The limit of f(x)^g(x) as x->0+ if f(x) & g(x) -> 0 from above must be ≤1. To see why, suppose f(x)^g(x)->a as x->0+. Then ln[f(x)^g(x)]->ln(a). Now ln[f(x)^g(x)]=g(x)ln[f(x)]. For x sufficiently close to zero, f(x)

[@victorsarmiento2552]

this is beautiful

[@shipsteer2891]

I know that this is from a long time ago, but wouldn't the 1/sqrt(x) be x raised to the power of 1/2 instead of -1/2?

[@asianhaydenxd]

No; the exponent is negative because it's in the denominator.

[@vcvishalchandra]

Thank you

[@tanushreebiswas9394]

Thanks sir..very nicely explained ❤️🤗

[@FairCheck]

Hi, I'm your big fan and I have one question that you could make a video... It's lim(x→0) (x^(1/x)) and lim(x→0) (x^(x / x))... Are these results same and is it possible to be solved. Thanks in advance

[@nightish_one6007]

Lim(x-->0) (x^(x/x)) should be easy, since you can just divide the x's and be left with lim(x-->0) (x^1),which is just x. Lim(x-->0) x^(1/x) can be solved very similary by using the second method bprp used ( writing it as lim(x--0) e^(ln(x^(1/x))) and then saying it is equal to e^(lim(x-->0) (ln(x^(1/x))), then calculating that limit), and I highly encourage you to try solving it yourself!

[@PlayGuy200]

Why did you have to differentiate when it was: lim x->0+ (ln x / (1/sqrt x)) ?

[@gregorsamsa9762]

L'hopitals rule, look it up

[@ayutac4056]

The nice thing is you can generalize this proof to show that x^f(x) -> 1 for x -> 0 for any superlogarithm function f :D EDIT: Ah wait no, f should be of the form x^a. f(x) = ln²(x) converges to 0 instead for example

[@bonzaiii3]

Great video!

[@libelldrian173]

I can't stop smiling. ❤

[@vincentlaw4580]

What is the limit for x>0-

[@justabunga1]

The limit DNE because it’s not in the domain due to the square root and the base cannot be negative.

[@juanjuan-mi4gi]

Como límite está bien....pero no tal símbolo cero a la cero es igual a uno...pues ello es solo un artilugio para justificar el cómputo tecnológico

[@thomvandenhil4717]

Why can't you just immediately enter 0⁺, so that you get: (0⁺)^√0⁺ = (0⁺)^0 = 1, since 0⁺ is like saying 0.0000001 and 0.0000001^0 = 1

[@MrConverse]

What’s the minimum of the function?

[@tennisiq75]

1/e^2

[@gabriellasso8808]

Lets define a^b: a^0 = 1 a^1 = a a^(m+n) = a^m a^n (Done for rational numbers) For a real number x, you can prove that for any Cauchy series x_n, a^x_n converges and define a^x to be the limit of that. So, by definition, 0^0=1. Why people still say it is undefined?

[@angelmendez-rivera351]

*Why people say still say it is undefined* That is a REALLY REALLY good question that nobody has been able to answer correctly.

[@isobar5857]

But why did you not just use the difference quotient, [[f[x] +[h]-f[x]]/h ? I'm lost !

[@shaidasabr5272]

Limit x apprach to zero for the square root of x is it exist?! i mean in general from both the left and right side if so can you explain

[@anshumanagrawal346]

No he said limit as x approaches 0 plus which means only the right hand side limit

[@derarken73]

Maybe im dumb but: Can't you look at the power first, say that sqrt(0+) approaches 0 and therefore every number to the 0 power is equal to 1?

[@a2333232332]

All most all f(x)^g(x) limit -> 1 when both f,g -> 0

[@_ssodaaa]

Can I use f(x) = x^x to consider ?????

[@blackpenredpen]

Yes. I did that before already. That's why I did x^sqrt(x) in this vid.

[@_ssodaaa]

@@blackpenredpen thanks. Very mush

[@theophonchana5025]

0^(0) = undefined

[@biscuit_6081]

Why does my calculator show 1 when I plug in 0^0?????????????

[@adhirachannel1148]

Can u plz help me how to integrate e^sinx

[@igxniisan6996]

I love you no humu

[@DadicekCz]

Or you could just instead of these long (but interesting!) calculations say this (For example with number 2) : 2^4 =16 2^3 = 8 2^2 = 4 2^1 = 2 So 2^0 must equal to 1, because everytime you divide the equation by 2 Works with all simple numbers

[@erikosterling3311]

no, that's just a way to say that x^0 = 1. The problem arrives when we realise that 0^x is always equal to 0. So if you have 0^0, then that's in the form x^0 so it ought to equal 1, but it's also in the form 0^x so it ought to equal 0. That's what indeterminate form means: you can't tell just by looking at the numbers what the result should be. 2*infinity is not indeterminate. x*infinity is always infinity and 2*x can be any number, including infinity. 2*infinity must therefore be infinity. No contradiction. 2*5 is not indeterminate. 2*x can be any number and 5*x can be any number. so 2*5 = 10 does not imply any contradiction. 0*infinity is indeterminate. 0*x = 0, |x*infinity| = infinity. So is |0*infinity| equal to 0 or infinity? Outside the limit world it's undefined, and in the limit world we need to look at the context. Lim x-->infinity of x * (1/x) will be equal to 1. Lim x-->infinity of x^2 * (1/x) will be equal to infinity. Lim x-->infinity of x * (1/x^2) goes to 0. Lim x-->infinity of (4+pi*i)x * (1/x) goes to 4+pi*i. They are all in the form 0*infinity but they all go to different values. That's what it means for a form to be indeterminate. If you want to you can try to find simple examples for each member in the indeterminate family (infinity - infinity, 0/0, 0^0, 1^infinity, infinity^0, infinity/infinity, 0*infinity) that shows they are indeterminate; two seperate limits in for example the form infinity - infinity that end up going to different values.

[@seroujghazarian6343]

@@erikosterling3311 you're taking LIMITS As far as non-limit numbers go, most of the above expressions are undefined, except 0⁰, which is 1

[@erikosterling3311]

​@@seroujghazarian6343 Let us assume 0⁰ = 1 and show that that leads to a contradiction. 0⁰ = 1 ln(0⁰) = ln(1) 0*ln(0) = 0 0*undefined = 0 If we assume 0⁰ to be 1 we get that 0*undefined is equal to 0. Undefined * 0 is undefined and as such 1 is undefined. However, 1 is defined, so there is the contradiction. If you thoroughly beleive that ln(0) is -∞ (which I don't much care if you do) then we still have a problem since 0*∞ is also undefined.

[@MuffinsAPlenty]

Erik Österling - Not a valid argument. Your argument also shows that (-1)² = 1 is wrong. Assume (-1)² = 1 ln((-1)²) = ln(1) 2*ln(-1) = 0 But ln(-1) is not 0, so we have a contradiction: two nonzero numbers multiplying together to give 0. The logarithm rule that ln(b^c) = c*ln(b) only holds in certain circumstances. You cannot apply it haphazardly.

[@angelmendez-rivera351]

@@erikosterling3311 *The problem arrives when we realize that 0^x is always equal to 0.* 0^x is obviously NOT equal to 0 always. For example, if x = -1, 0^x is not 0. If x = i, 0^x is not 0. There is no reason why 0^x should be 0 if x = 0. *So if you have 0^0, then that's in the form x^0 so it ought to equal 1, but it's also in the form 0^x so it ought to equal 0.* This is an invalid argument, because while being of the form x^0 "should imply" equality to 1, I already explained above that 0^x is not always equal to 0, and there is no reason to assume a priori 0^x should be 0 for a given value of x. *That's what indeterminate form means: you can't tell just by looking at the numbers what the result should be.* No, this is definitely not what "indeterminate form" refers to. Indeterminate forms are, by definition, limit expressions, which under certain restrictions, cannot have their value deduced from those restrictions alone. 0^0 is an arithmetic expression, and so, it is not indeterminate. *x·infinity is always infinity* Ah, so we are just going to ignore the existence of x = 0. I see. Anyway, this is technically nonsense, since ♾ is not a number and you cannot do multiplication with it. *0·infinity is indeterminate.* No, 0·infinity is nonsense, because "infinity" is not a number. lim f(t)·g(t) (t -> c), with the restrictions that lim f(t) (t -> c) = lim 1/g(t) (t -> c) = 0, is indeterminate. This is not the same thing as the expression "0·infinity." *Let us assume 0^0 = 1 and show that it leads to a contradiction.* It does not lead to any contradictions whatsoever. *0^0 = 1; ln(0^0) = ln(1); 0·ln(0) = 0* ln(x^y) = y·ln(x) is not true in general. (-1)^2 = 1 implies ln[(-1)^2] = ln(1). If we assume that ln(x^y) = y·ln(x) is true in general, then ln[(-1)^2] = 2·ln(-1) = ln(1) = 0, so 2·ln(-1) = 0, which is clearly false. Does this mean (-1)^2 is undefined? No. (-1)^2 = (-1)·(-1) = 1 is a fact that can be proven from the field axioms. What led to the contradiction was claiming that ln[(-1)^2] = 2·ln(-1) is true, when in fact, it is false, and ln(x^y) = y·ln(x) is false in general. It only holds for rather special cases of x and y. So your proof is invalid, and it fails to demonstrate that 0^0 = 1 leads to a contradiction. This is consistent with my claim above: there are no contradictions that you can prove with this "assumption," which is in reality a direct consequence of the definition of exponentiation. When n and m are natural numbers, n^m denotes the product of the elements of the m-tuple where all its elements are equal to n. Therefore, 0^0, by definition, denotes the product of the elements of the 0-tuple where all its elements are 0. Since the 0-tuple is the empty set, the product of its elements is vacuously 1, because the empty set has no elements. Therefore, 0^0 = 1. Actually, the secret in the identity x^0 = 1 is the fact that the 0-tuple contains no elements, so it does not matter if the base is 0 or not: the base is never contained in the 0-tuple, which means the 0-tuple is unique and base-independent. You can prove the 0-tuple is equal to the empty set, which is also unique, by the way. So the only way you can dispute 0^0 = 1 is if you dispute that the product of no elements of the empty set is 1. Good luck disputing that, and if you dispute that, you would be proving x^0 = 1 is false for all x, not just x = 0.

[@santhandevid6378]

My school doesn’t allow to use L’hopital

[@justabunga1]

You have to use Hopital’s rule because there is no way you can use algebraic method for this. If the AP exam or university exam is coming up (and if no graphing calculators are allowed), you have no choice but to use this method.

[@kamalrihani9609]

by definition lim xlogx =0 when x reach 0

[@sabergad9265]

Good but lighting reflected on w borad

[@rhc1560]

But in a video u said 0^0= 1 by convention

[@Nickesponja]

Is it the case that lim(x->0+) x^(x^r)=1 for all r?

[@Lucas-zd8hl]

With a number that close to 0, and any exponents not equal to 1 or 0, I would round it off to 0

[@marlinjarms5992]

Well i used desmos. If r>0 it looks Mike your statement is true. If r=0 then x^(x^0)=x If r0+) x^(x^r)=0 I know it is not a prove but it looks Mike your statement is true as long as r>0

[@dwaraganathanrengasamy6169]

Well,... Its time for shortcut. We very well know lim y->0 y^y = 1. The question is, limx->0 x^√x = Limx->0 ((√x)^√x)² = 1² = 1...! Tricks in limits(use of previous knowledge) is a lot enjoyable. I love limits😋💖💖💖

[@pon1]

If it's infinitely approaching 0 can't it be said to be exactly 0 then? Just as infinitely approaching 1 like a third times three is exactly 1 even if it is 0.999999... (infinitely approaching 1)? So that 0^0 is 1 just as any other number ^0.

[@maciejlabanowicz8640]

I see there is an undefined blue pen.

[@bounpakhongphoomithep1640]

2³ = 1x2x2x2 2² = 1x2x2 2¹ = 1x2 2º = 1 = 0º

[@orisphera]

I think the second problem can be reduced to lim(x->0)x**x

[@MikehMike01]

According to Wikipedia 0^0 = 1

[@BizVlogs]

Guys help me. Does the (limit of the) indeterminate form 0^0 ever equal something besides 1? Can anyone give me an example? (Don’t say 0^x because that’s not really indeterminate form)

[@MuffinsAPlenty]

You can take things of the form (e^(-x^2))^(-c/x^2) where c is any real number. Then take the limit as x approaches infinity. The base, e^(-x^2), tends to 0. The exponent, c/x^2, also tends to 0. So we have the limiting form (→0)^(→0). However, using exponential rules, for all sufficiently large x, we have e^(-x^2*-c/x^2) = e^c. So the limit is e^c. (Again, c can be any real number, so the limit could be any positive number.) But there is a theorem that if your two functions (the base function and the exponent function) are sufficiently nice (they have power series expansions defined around 0), then you will always get 1 for the limit. So yeah, most of the examples people come up with for (→0)^(→0) will ultimately have a limit of 1, since they tend to stick to nice functions. But it is possible to get something else.

[@sk4lman]

That derivation disturbs me a bit. Shouldn't the value you get in the end really be the derivative as x->0?

[@angelmendez-rivera351]

sk4lman Why should it?

[@sk4lman]

@@angelmendez-rivera351 It seems to me that once he differentiates he's describing the properties of the derivative, and not the original function.

[@angelmendez-rivera351]

sk4lman He did not differentiate the function though, he used L'Hopital's rule.

[@lakshmivalavan2389]

i think it is not if it is wrong please don't bother my answer answer: if you take any number like 3^3=27 3^2=9 3^1=3 3^0=1 why?; take;3^0 we are doing multiplecation zero times is 1 i mean we are not multipleing it [ 1*3^0 = 1] as the muliplecative identity dosen't change the value the 3^0 is just nothing and i don't mean about 0 i mean we are not doing it like; 1*3^0=1(no operation have made here)=1 now if we replace 3 as 0 the same works and the most told idea about 0^0 =0^1 * 0^-1 where it is told to not take resibrocal of 0 1 0^0 -- = -------=0^-1 where it dosen't make any sence that we can't abel to assign two value to it thats why we can't take 0^-1 0 0^1 ad limits approches the value and non of the calculas can give a accurate value

[@studentkaschool6902]

Nice

[@lazyperson7343]

0⁰=0¹‐¹=0/0, which is undefined

[@angelmendez-rivera351]

No. It is not true that 0^(1 - 1) = 0^1·0^(-1). 0/0 is undefined, but 0^0 = 1. a^(m + n) = a^m·a^n only if a^(m + n), a^m, a^n all exist.

[@lazyperson7343]

@@angelmendez-rivera351 1) your comment doesn't defy my comment in no way and 2) It is a definition, that means that it is not possible, i just tried to interpret it in a clever way

[@angelmendez-rivera351]

@@lazyperson7343 *your comment doesn't defy my comment in no way* No, it definitely does. If you think it does not, then you do not understand your own comment, or you do not understand my comment, or both. *It is a definition, that means that it is not possible* The definition of exponentiation directly implies 0^0 = 1. There is nothing you can say to disprove this. If you are saying that a definition is not possible, then you do not understand what a definition is. *I just tried to interpret it in a clever way.* You tried, but you failed, because the argument is not clever, since it is not even a valid argument.

[@lazyperson7343]

@@angelmendez-rivera351 By being aggressive and hostile towards me does not allow a fruitful discussion to occur. By my comment I implied that a definition in Math is something that can not be proved, which means that it can't be disproved either. Your comment has no correlation with 0⁰ being equal to 1, because m and n should be n and -n in order for it to be equal to 0, but raising 0 to a negative number is not possible because you cannot divide by zero. I hope you find my comment elaborate because I'm not a native English speaker so I guess that's where the misunderstanding comes from my second or first comment

[@angelmendez-rivera351]

@@lazyperson7343 *By my comment, I implied that a definition in Math is something that can not be proved, which means it can't be disproved either.* Okay, this is correct, but while the definition itself cannot be proven or disproven, implications of the definition can be proven or disproven. *Your comment has no correlation with 0^0 being equal to 1, because m and n should be n and -n in order for it to be equal to 0* My comment does have correlation here. My comment stated that a^(m + n) = a^m·a^n can only be true if a^(m + n), a^m, a^n are all defined. This is true for arbitrary n and m, and so it follows trivially that it is obviously true when m = -n. As such, if a^(-n) is not defined, that tells you nothing about whether a^n or a^0 are defined or not, because a^(-n) not being defined means the equation a^0 = a^n·a^(-n) is not true anyway. Of course, this does not prove 0^0 = 1, but it does show that your argument is invalid, and so you cannot use it to show that 0^0 is undefined. However, you can indeed prove 0^0 = 1 by simply using the definition of exponentiation. For natural n and m, n^m denotes the product of the elements of the m-tuple where every element it contains is equal to n. Therefore, 0^0 denotes the product of the 0-tuple where every element it contains is equal to 0. However, the 0-tuple contains no elements, and the product of no elements is 1. Therefore, 0^0 = 1. In fact, this argument proves x^0 = 1 for every x, because, as the 0-tuple contains no elements, it is independent of the base x, and the product of no elements is also independent of the base x. The only way yo can disprove 0^0 = 1 is if you disprove that the product of no elements is 1. However, if you do this, then you will necessarily be proving that x^0 = 1 is false for every x, not just false for x = 0. Of course, you cannot disprove that the product of no elements is 1. If you apply an operation that is commutative and associative and has an identity element to the empty set, then the output necessarily has to be said identity element.

[@HenrySchach]

In comparison, lim x-> 0+ x^x is 1

[@YodaWhat]

blackpenredpen -- I wish you would talk a little more slowly. You have bursts of high-speed speech in which I cannot understand a word you say... and I understand Isaac Arthur perfectly.

[@comingshoon2717]

Me impresiona