No joke I was suffocating for those few seconds when he went forward without that negative. Just shouting at my laptop to somehow make that negative sign appear out of somewhere. Guess it worked
not enough people talk about how well you manage multiple markers in one hand. The way you cleanly switch between colors is really cool to just watch because the math goes way above my head 😅
same. and i had a pretty good idea of what the final form was going to look like and was kind of looking forward to him getting to the end and finding that 0^0 = infinity
I'm used to BPRP being clever, and very smooth with proofs. But this is the first time I've seen the man so _aggressively_ math. It's scary but in a comforting way.
I love how you can share your findings not just in a random paper published to some journal, but on youtube! It's stuff like this that reminds me how much I love mathematics, and your channel... :D
I can’t believe it. I’ve watched the video twice and done the calculations along with the video both times, and the math checks out. I’m both pissed off, and extremely impressed well done. Well done indeed. Have a Merry Christmas, and a wonderful New Year.
This will now serve as a great example not only of your example mathematically but of how a subject that can be mundane and boring or disinteresting, such as mathematics and limits and derivation, can become incredibly engaging when given the right individual presenting it. It also, specific to me, will serve as further proof that I'm a nerd, bc I just sat here thrilled watching you do limits and understood every step of it, not knowing about the significance of this concept nor the purpose in the example, but simply loving the mathematical process you went through. This is how I have fun.
This reminds me of one mathematician in the 19th century who used the bizarre notation 0^0^x. He said that when x is positive, 0^x=0, so 0^0^x=0^0=1. When x=0, we get 0^1=0. When x is negative, 0^x is infinite, so 0^0^x=0 again. Therefore, 0^0^x is the function that is 1 when x is positive and 0 when x ≤ 0. EDIT: It's true that 0^0 and 0^(negative number) don't make sense mathematically. I'm just repeating Libri's argument here. For more about this, Donald Knuth has an interesting paper called "Two Notes on Notation" that mentions this story.
1:10 Admit it. When he started getting emotional, you full-on did that reflexive, empathetic gasp of response at his emotion. I'm still trying to recover. Math is so beautiful. 😭
@@angeldude101 0^0 is always 1 . But the _limit form_ 0^0 is an indeterminate form. Likewise, 1^infinity is always 1 ; but the _limit form_ 1^infinity is an indeterminate form.
In my country, instead of writing the limit as 𝑥 → 0⁺, we write the limit as 𝑥 ↓ 0 and instead of writing the limit as 𝑥 → 0⁻ , we write the limit as 𝑥 ↑ 0. :)
This has literally helped me better understand limits fundamentally after 12 months doing calc courses combined. A really bad 12 months where i learned a lot about failure, but still! wow!! What a pretty solution
I saw the thumbnail and I was filled with rage and confusion. But once I saw your function, I realized I was about to be wrong. The big 'oh shit' moment for me was at 11:52. I actually gasped. Very nice function!
If you read the wikipedia article for 0^0, it gives a bunch of examples for limits of the indeterminate form 0^0, but they all approach different values. For example, lim x to 0+ of (e^(-1/x^2))^x approaches 0, but lim x to 0+ of (e^(-1/x^2))^-x approaches -infinity. The limit lim x to 0+ of (e^(-1/x))^(ax) seems to always approach e^-a, which is not a constant value like 0. So you can't actually find a limit that gives the "correct" value as it approaches 0^0.
This is another example of the definitional difference between something that approaches zero in the limit and zero itself. Various sums that approach zero in the limit will give various values of the limit of "0^0" while strictly 0^0 remains undefined, so there is no "correct" value to it. On the flip side (taking the inverse) of this is the fact that infinity exists outside the real numbers, so various sums approach infinity in the limit but they do not strictly equal infinity.
Yes, that's why it's called an indeterminate form. The same is true of others like 1^inf, 0^inf, 0/0, inf/inf etc. The answers depend on the limit functions you take to get there. This is what defines an indeterminate form. The purpose of this video is not to show that 0^0 equals anything, but rather that it *can* equal 0 if you set the limiting equations up correctly. I do feel like that should've been made clearer in the video. Edit: as pointed out below I made a mistake in saying that 0^inf is an indeterminate form
@@alansmithee419 0 to the power of infinity is not indeterminate. However, infinity to the power of 0 is. Also, indeterminate forms yield Aleph-Null as the answer, as we don't know the cardinalities, and also, the answer can be any number in an interval. Indeterminate forms are created because of you are trying to undo an "annihilation" function. An annihilation function yields only one output for all of its inputs, so if an inverse exists, it will have one input but have infinity outputs. However, on any occasion, only one answer can be correct, but because we don't know the cardinalities, all numbers within the interval is vacuously true, as a vacuous truth is defined as if a prerequisite is required to determine the truth or falsity of something, and that prerequisite is not present, we are unsure if it is true, so we will consider it as a vacuously true statement. Therefore, we can consider 0 divided by 0 to be equal to Aleph-Null, with all elements in that set to be vacuously truly equal.
I don't follow your channel, and I don't even have to do much math in my everyday job or life. But this legit made me miss calculus for the first time in 15 years. How it felt so much like the art of being clever. This is a beautiful proof.
How can you take the natural log of that limit if it equals to 0? Isn't ln(0) undefined? Isnt that a contradiction in your proof? Or am i missing something here?
Agree, seems that you start with a non-defined operation (Ln(L) is valid only for L > 0, right?). So if your result is L = 0, you start with a contradiction (it seems).
Saw it on desmos from 10^199 to 10^200, the ln(x) function is decreasing but still far from 0 (0.1631), and the square root function is near to 0. Wow! Thanks!
I get the emotional feelings behind this video because learning about some cool math thing that you thought wasn't possible or was really difficult is an emotional experience
Wtf almost always 1? if you take ln both sides and assuming the 0 on the bottom is never negative then lnL=0.ln0=0.-inf=-0.inf, so every 0.inf limit that is not 0 is counter example because e^m /= 1 if m /= 0
@@rafiihsanalfathin9479idk what you’re saying for a lot of the comment, but what the commenter is saying is that most limits that when plugged in give 0^0 are equivalent to 1. If you take a class that involves L’ Hopital’s rule then you will probably notice this. It doesn’t mean that 0^0 is always equal to one, just that it does for many limits
@@kart338_QK what im saying is that limit that have the form of 0.∞ but have the value other than 0 counter example of what the commenter said. For example lim x->∞ 1/x . -x = -1 (ik this is crappy example but whatever), we can write -x into ln(e^-x) then we got lim x->∞ ln(e^-x)/x=-1 so lim x->∞ (e^-x)^(1/x)=1/e. In general any limit that have the form 0.∞ with the value other than 0 is a counter
14:32 "this right here, it's like the biggest zero.. But! If i have 1/ln of x, that was still not enough. So.. i put another one, and it worked!" Great idea, brilliant solution, deserved joy of Eureka, deserved like ❤
@@lolerie Maybe it is shocking to you... When (an) is any sequence convergent to 0+, obviously the sequence (an)^(-1/ln(an)) tends to 1/e. It follows that - if (bn) is a sequence that goes to 0+ significantly faster than -1/(ln(an)), then (an)^(bn) goes to 1, - if it goes to 0+ significantly slower than -1/(ln(an)), then (an)^(bn) goes to 0. And obviously the limit can be made equal to anything, it's just a matter of how (bn) compares to (-1/ln(an)).
By assigning L := lim(...), it acquires a fixed value (which you hypothesize to be 0). In that case, taking ln(L) is invalid, because ln is not defined at 0. On a separate note: have you tried visualizing x^y in 3D space? It might give a visual intuition at least. I'd be curious to see a multi-variable limit calculation of z = x^y, x->0, y->0.
This is only a problem in the sense that it highlights the difference between a limit approaching zero and being equal to zero. By setting L :=, he is not saying L is literally ' 'equal to' but that the value of L is assigned the value of the the approachment. Recall, that the definition of a limit doesn't assign a value to the limit. In this case, for all epsilon > 0, there exists a delta > 0 such that ... L < epsilon.
@@rajeevram4681 what? of course limits have values; that's the whole point. otherwise integrals wouldn't have values. the expression on the inside can be said to approach the limit, but the entire point of the lim operator is to evaluate that limit
May I suggest purchasing refillable dry erase markers? Perhaps, if I may be so bold, one black and one red? They write much nicer and more consistently. They are cheaper in the long run for someone who uses whiteboards often. They are better for the environment. The nibs are replaceable as well. I got some that are made by Pilot. They're amazing. Edit: I see you used a blue one in there, so go for it! You earned it with this proof.
I prefer the argument for 0^0 being 1. Consider f(x) = x^x. f'(x) = x^x (lnx+1). Roughly: We examine lim(x→0+) of f(x). We can see that the sign of f' near 0 is < 0: Let D (delta) be positive. If D is small enough, ln(D) < -1, ie, ln(x) < -1. So ln(x)+1 < 0.Then it's also true that x^x (ln(x)+1) < 0. Since f' is negative for small enough D, f(x) is finite increasing as x approaches 0 from the right. And as it does, f(x) gets closer and closer to 1. So f(x) has a definite limit, which, I submit, is 1.
Why are you examining the function x^x, and not x^y? It's not like base and power should always be equal to each other. Sure, if the only case where you use powers satisfies this, then this argument works. But in most cases this restriction is too strong, so you need to look at function of 2 arguments f(x, y) = x^y.
The usual way to make 0⁰ approach any positive number C (at least the way I usually do it) is to take the limit of (e^(−1/|x|))^(−ln(C)×|x|) as x→0. Maybe this is not a good example in that the expression immediately simplifies to C, so there's no real work in taking the limit, although at least neither the base nor the exponent is constant this way. But of course it doesn't work for C=0.
Finally, I can be watch a daily upload! Btw I wanna say that you are my hero. Because of you I have found my love for math and am commited to going into theoretical physics. Thank you. ❤
Well done, especially with the stage walk-off at the end (mike drop!). On the one hand, 0^0 can be any non-negative number, so one can say that 0^0 is undefined. On the other hand, 0^0 can be defined to equal 1. This definition makes the most sense, since it removes the discontinuity in functions like x^0.
Limx->inf (sqrt(2x+1)-sqrt(x))=Limx->inf((2x+1-x)/(sqrt(2x+1)+sqrt(x))=Limx->inf((x+1)/(sqrt(2x+1)+sqrt(x))=inf because a linear function grows faster than a sqrt function
Let a=1/2 and x>0. (2x+1)^a-x^a > (2x)^a-x^a = (2^a-1)x^a. Since the right side goes to infinity as x grows, so does the left side. Also, a simpler example is the limit at zero of f^g where f(x)=e^(-1/x^4) and g(x)=x^2.
@@opensocietyenjoyer That doesn't work. In order for f(x) to approach 0, you need x approaching negative infinity. However, you can't have x approach negative infinity when talking about x^(-1/2).
lim((1/(e^x))^(1/ln x)) also goes to 0 as x approaches infinity. And lim((1/x)^(1/ln(ln x))) do that too. And so on. Just get some very slow decreasing function for exponent and very fast decreasing one for base.
Hey BPRP =) Considering how many people noticed the missing minus symbol, you are obviously doing a great job. Your presentation is well organised so it is easy to spot a mistake and your viewers are well enough educated to understand the error. Your are awesome and this fake proof looks very convincing :D I will have my students try to spot the false assumption ^^
Thank you for the nice words! However, I am not sure what you mean by "fake proof". This video isn't about "show 0^0 equals 0", it is about "a limit with the indeterminate form 0^0 being 0". You can also check out my other videos that 0^0->1 and 0^0->e. Cheers!! : )
10:28 I am just curious... did you drop the negative? Shouldn't it be -xlnx ? Since you multiplied by -1 to cancel.. and the cancel would result in -1 in the right numerator.. ? EDIT - he fixed it. Thank god.
I'm probably missing a point here, but I think there's a simpler way: lim x->infinity of ( (x^(-x))^(1/x) ) both x^(-x) and 1/x approach 0 as x approaches infinity (x^(-x))^(1/x) simplifies to 1/x which is already known to approach 0.
Why don't you use epsilon-delta proof to show this limit is 0. But this example is very nice. It's necessary, but not sufficient. Actually, I take that last sentence back - these are two different functions. And convergence is a property of the function. Even if their behavior seems the same at the point, it does not mean if one converged so will the other. You'd have to show that somehow there is an upper and lower bound error that converges to zero. Do this proof. Then if checks out, you might have a claim. But by the time you do that, why not go back to epsilon-delta proof.
i have a doubt about the premise of the problem. If x->+inf everything works nice, but 0 as a number can both be reached with a positive limit and with a negative limit. If you plug in -inf in the limits, it doesn't work. I just didn't quite understand this.
@@enderforces7013 Which limit? The limit of [f(x)]^g(x) for these particular functions f(x) and g(x) as x goes to +infinity _is_ defined, namely it is 0 . Just as, for example, the limit of (1/2)^x exists for x --> +infinity , even though it doesn't exist for x --> -infinity. But you may have a point: can we find functions f(x) and g(x) such that the limit of [f(x)]^g(x) is 0 when f(x) and g(x) simultaneously approach 0 ; not only when f(x) and g(x) approach 0 from the positive side but also when f(x) and g(x) approach 0 from the negative side? blackpenredpen, your job is not yet done!
We can generally define that 0^0 would be equal to 1 or 0, but in my opinion define that is 1 or 0 is not true. As how can 0 transform to 1 or any other number, if is lower? So it's generally can be defined as it's result: 0^0 = Undefined. (You can't turn 0 to itself if you're powering it to a higher number, so that's why is undefined)
Hi, I have a mathematical question. I'd be happy if someone will help me with it. If you use Euler's identity, you can see that e^(iπ) = -1. Now, square both sides to get e^(2iπ) = 1. Now take the natural log on both sides, and 2iπ = 0. And now, divide by 2i to get π = 0. How is this working?
im not even taking calculus yet but my guess is that ln only takes the principal value of it because with imaginary numbers exp function is cos + isin its like how 0 is not the same as 2pi just because they have the same cos value
ln(e^2ipi) is not the same ln as ln(1) (I THINK. IM NOT AN EXPERT TAKE THIS WITH A GRAIN OF SALT). ln can be treated as the inverse of e^x when dealing with complex and imaginary values and not a simple log function, so you are not performing the same operation to both sides of the equation I don’t think. Again, this is almost certainly inaccurate somewhere considering I’m not a mathematician.
Credit to Akiva Weinberger "On the complex numbers, the logarithm isn't a function; rather, it's a multifunction (returns multiple values for one argument). This is how e^(2πi)=e^(0) doesn't imply 2πi=0 after taking logs; ln(1) is all integer multiples of 2πi"
In complex world we dont use just ln, we use Ln (starting from the capital letter). They’re quite similar, but Ln produces infinite amount of outputs for one input Actually, there are more functions in complex analysis which are analogous to normal ones and they are distinguished by that capital letter
Remember that Euler's formula tells us that e^(iθ) = cosθ+i.sinθ. So, when we evaluate e^(i2π), we get cos(2π)+i.sin(2π), which gives us 1+0=1. But we also have e^(i2kπ) = cos(2kπ)+i.sin(2kπ) =1 for k ϵ Z. Because Cosine and SIne are cyclic with period of 2π, any "inverse" of them will not be a function. Recall that invertible functions must be 1-1 and onto. So, we can't really have a usual kind of inverse of exponentiation (logarithm) when using complex powers. The best you can do is recognize that seeking the inverse of a complex exponential will generate an infinite set of solutions of the form a+i.(b+2kπ) for k ϵ Z and a,b ϵ R. As noted by @user-yy7bq1zx8r, this Complex Logarithm is notated using a capital L (Ln or Log). Have a look at the Wikipedia article on Complex Logarithm to start digging deeper. en.wikipedia.org/wiki/Complex_logarithm
We can only seperate the limit if both limits exist. In this case since lim(lnx) goes to infinity as x goes to infinity, the limit does not exist. So the seperation does not work here. (Or am I missing something?)
A legitimate concern! Here's a video I found that explains it: th-cam.com/video/H2RQC4PxEM0/w-d-xo.html ("When can we switch the limit and function?" by Mu Prime Math).
I'm in the obligation of congratulating you for the massive amount of effort put on this video and solve one of if not the most confusing undeterminations in math. Amazing work and awesome video!!!🎉🎉🎉
@@fabiod.674 this function doesn't represent itself but a group of functions like a archetype of functions (you can add a infinite amount of constants in a couple of places and will be limited by 0 anyways) and what it proves is that 0^0 is in fact limited by 0 in some cases (this kind of cases).
Even with all the effort you took to get that thing to approach 0 it is worth noting that it is an incredible slow function. It actually has a positive slope until it hits around 50, and after that it just goes down glacially slow. The y value is still at 0.005 when x is at 10^15. That is insane for a function that approaches 0, especially when you consider that that hump only peaks at around 0.143.
@@jjjannesthe limit does actually go to 0 though, albeit very slowly. Could you set L(x)=sq(x+1)-sq(x)..., then do ln on both sides, then take the limit?
As a long time viewer since before your channel became so popular, I love to how passionate you were working this out! I love exploring exponents of zero myself and was in the middle of writing up an idea for working with exponents equal to or less than zero. I stopped working on it after dropping out of a mathematical physics bachelors degree during the height of covid isolation and my poor mental health. After seeing this video I have to ask, would you be interested in talking to me about it?
wow I learned so many math rules in just 15 minutes! the hospital rule is strong. thank you for your demonstration. I hope you can read this, from a French math enthusiast
The confusing here is that we aren't actually evaluating 0⁰. We are evaluating the limit, L, of the function f(x)^[g(x)], where f(x) and g(x) approach 0, as x approaches 0. We aren't saying 0⁰ approaches one particular value and based on the choice of f(x) and g(x) the limit L appear to be able to take any value we want it to.
I looooooove the way that you were explaining. That was so cool. With that negative, I was just dying 😅. And I totally understood your feelings when you finished it.
imo calculus is just formulas; make sure you are good with algebra and a little bit of trig (understanding trig identities and understanding unit circle)
So the problem you faced for so long, and have at last solved, was to find f and g such that the limits as x-> 0⁺ of f(x) and g(x) are both 0, while that of f(x)^g(x) is also 0. Congratulations! It seems like there should be a simpler solution, but perhaps there isn't. Fred
@@blackpenredpen Yes, and rightfully so. Meanwhile, I'm trying my hand at other solutions. BTW, I misstated the problem in my comment. Should have said "... limits as x->∞ ..." I think they are essentially equivalent, though, by simply replacing the argument of f and g (i.e., x) with its reciprocal, 1/x.
By using calculator to estimate the limit 0^0: 0.00000000001^0.00000000001=0.999999999 (very close to 1) -0.00000000001^-0.00000000001=-1 As a result, I suspect limit of x^x (x -> 0) doesn't exist
I was about to write a comment about the negative sign but I decided to see the result first because it might not be that important. He put it back in and my heart went back to its place.
It feels like we missed something. When we use so many tricks it is easy to make a small mistake. My first guess would be that we played too much with infinity for this to work. But I'm not sure... Maybe we could start a discussion about this, or maybe you can make a video going in depth about what could be wrong with this derivation? Thanks for a cool video tho!
Yeah, take a look at 2:18. If you carry out the multiplication on the numerator, this is X+1-X = 1 identically. But not so if you look for the values of the polynomial. For example (1+1)^1/2-(1)^1/2 is not equal to 1. Neither is (1+1)^1/2+(1)^1/2. I think it’s important that the square root of a polynomial is not a polynomial. We could convert the sq roots to a polynomials to actual polynomials with a Taylor series but it might not be necessary. If you plot [(x+1)^1/2 - (x)^1/2] and [(x+1)^1/2 + (x)^1/2 you get two functions that cross, form a node. So you don’t get a single value. It is indeterminate. I think this is explained better with some basic concepts in algebraic geometry, but I would have to review it. But yes, I think this derivation of 0^0 might be more complicated.
@@michaelzumpano7318not sure if I understood your point correctly, but he's just using the special product formula: (a + b) * (a - b) = a^2 - b^2 Thus: (sqrt(x + 1) + sqrt(x)) * (sqrt(x+1) - sqrt(x)) = sqrt(x+1)^2 - sqrt(x)^2 = x + 1 - x = 1. It's not that complicated…
@@light_asaii4858 I'm sure in some cases you can do that. In most cases actually. Actually, only if the fraction can be simplifed to 1, then you take the limit after, and because there is no 'x' left, the limit become what is left, which is 1 in this case.
I was about to complain he left out the (-1) from what he factored out, until I watched the video to the end. The lesson of today, be patient to the end before posting a comment. Over all, awesome video, awesome explanation, some what easy step to follow.😎👍🏽
man, BPRP always delivers.
Awesome video, very clever idea and reasoning.
Here’s the man!!!! Btw I always remembered that comment and I was like wow finally!!!’
@@blackpenredpen I am flattered
it should take you a minute to find a much simpler example:
f(x) = e^x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
i think yoou mean e^-x because e to the x goes to infinity and when x goes to 0 g(x) goes to infinity@@opensocietyenjoyer
😯😯😯😯
This limit should appear in wikipedia as "blackpenredpen's limit".
Oh yeeeeeh
he also used blue pen you know.
@@tintiniitk in the information table: Pens used are black pen, red pen and blue pen (?).
On it!
(Na, I'm joking. Would be great if someone did it though!)
Agree
No joke I was suffocating for those few seconds when he went forward without that negative.
Just shouting at my laptop to somehow make that negative sign appear out of somewhere. Guess it worked
lol thanks!!
I would bet he would have re recored whole thing lol
Infinity is not a defined number. I think there are flaws in his assumptions.
@@sparxumlilo4003 Nor is Pi.
The negative sign, e to the infinity is zero, not caring about ln, the ending... so many great tension moments. A big thumb up!
🤣
it should take you a minute to find a much simpler example:
f(x) = e^x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
Thank u!!!
not enough people talk about how well you manage multiple markers in one hand. The way you cleanly switch between colors is really cool to just watch because the math goes way above my head 😅
Thank you!!
He's like a live printer
When you get familiar to using chopsticks, that would be easy.
it's pretty impressive 👍
He is very proficient in that skill
I almost died with the negative sign.
same. and i had a pretty good idea of what the final form was going to look like and was kind of looking forward to him getting to the end and finding that 0^0 = infinity
I was stressing a lot 😂😂😂😂
I am sorry…
I was in pain
SAME I WAS SO CONFUSED
So basically, 0^0 approaches 0 when the base approaches 0 much, much quicker than the exponent.
х^х вообще то к 1 приближается при х->0
@@СвободныйМатематикlim f(x)^g(x) could be any value, when f(x)->0 and g(x)->0
Well done on finding a legitimate form where it does approach zero. *_And it worked!!!_*
it's very easy to find a much simpler example:
f(x) = exp(-x) → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
@@opensocietyenjoyer as x goes to infinity of course
lim does not commute with this mapping.
@@opensocietyenjoyer I dont think you understood the exercise at hand.
@@ciarangale4738 i did.
I'm used to BPRP being clever, and very smooth with proofs. But this is the first time I've seen the man so _aggressively_ math. It's scary but in a comforting way.
I love how you can share your findings not just in a random paper published to some journal, but on youtube! It's stuff like this that reminds me how much I love mathematics, and your channel... :D
Thank you!!
your feelings are irrational
I can’t believe it. I’ve watched the video twice and done the calculations along with the video both times, and the math checks out. I’m both pissed off, and extremely impressed well done. Well done indeed. Have a Merry Christmas, and a wonderful New Year.
This will now serve as a great example not only of your example mathematically but of how a subject that can be mundane and boring or disinteresting, such as mathematics and limits and derivation, can become incredibly engaging when given the right individual presenting it. It also, specific to me, will serve as further proof that I'm a nerd, bc I just sat here thrilled watching you do limits and understood every step of it, not knowing about the significance of this concept nor the purpose in the example, but simply loving the mathematical process you went through. This is how I have fun.
I love that in your search for this solution, you were looking for "the biggest zero"
Thanks 😆
your feelings are irrational
This reminds me of one mathematician in the 19th century who used the bizarre notation 0^0^x. He said that when x is positive, 0^x=0, so 0^0^x=0^0=1. When x=0, we get 0^1=0. When x is negative, 0^x is infinite, so 0^0^x=0 again. Therefore, 0^0^x is the function that is 1 when x is positive and 0 when x ≤ 0.
EDIT: It's true that 0^0 and 0^(negative number) don't make sense mathematically. I'm just repeating Libri's argument here. For more about this, Donald Knuth has an interesting paper called "Two Notes on Notation" that mentions this story.
Based!
The proofs you gave are just red herrings for arbitrarily setting 0^0 = 1 and 0^inf = 0.
Ah! So in essence we have f(x) = 0^0^x as a mathematical notation for a _step function_ (which is a primitive of the Dirac delta function).
@@laurentmeesseman4286 I'm not claiming those equations are valid - I'm just giving the original rationale for that notation.
@@laurentmeesseman42860^0 equaling 1 isn't arbitrary
1:10 Admit it. When he started getting emotional, you full-on did that reflexive, empathetic gasp of response at his emotion. I'm still trying to recover. Math is so beautiful. 😭
Your smile while revealing key steps throughout the whole video made my day! 😄
Im sorry, I'm still not happy with this. Your name is "blackpenredpen", and you did not use a black pen and red pen. Please redo this.
Counterpoint: pause 4:47 and look at the board.
Did you watch the video? Maybe you should lol
You guys they mean PEN. he’s using expo markers 😂
@@-.SkyArt.- expo markers are a type of pen, you just dont know your definitions
@@-.SkyArt.- ball point isnt the only type of pen
This was a must watch. Thank you for reminding me 0^0 is not just almost one
This limit form is almost always 1.
@@lolerie Keyword: "almost." There's a reason it's considered an indeterminate form.
@@angeldude101 0^0 is always 1 . But the _limit form_ 0^0 is an indeterminate form.
Likewise, 1^infinity is always 1 ; but the _limit form_ 1^infinity is an indeterminate form.
@@yurenchu 0^0 and 1^ infinity make no sense mathematically unless you're talking about the limit forms. Or would you argue that 0/0 is also always 1?
@@Felixr2Shouldn't 0^0 be 0 since your basically multiplying 0 by itself?
14:40 that mic drop was epic
SPIKED it like scoring a touchdown.
DAMN!😮
PEN SLAM! (C) FIFIWOOF 2023 ALL RIGHTS RESERVED
14:45 DAMN!!!!!!
I'm SO in love with you right now BlackPenRedPen!
DAMN!!!!!
I waited 6 years for this! This is great!!
it should take you a minute to find a much simpler example:
f(x) = e^x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
Thank you!!
In my country, instead of writing the limit as 𝑥 → 0⁺, we write the limit as 𝑥 ↓ 0 and instead of writing the limit as 𝑥 → 0⁻ , we write the limit as 𝑥 ↑ 0. :)
that makes way more intuitive sense!
I'm gonna use this from now!
we write x->0+0 and x->0-0
What country
@@ightimmaheadout290netherlands
This has literally helped me better understand limits fundamentally after 12 months doing calc courses combined. A really bad 12 months where i learned a lot about failure, but still! wow!! What a pretty solution
I saw the thumbnail and I was filled with rage and confusion. But once I saw your function, I realized I was about to be wrong.
The big 'oh shit' moment for me was at 11:52. I actually gasped. Very nice function!
Your enthusiasm earned a subscriber.
Please don't lose that love and fire for what you do! :)
If you read the wikipedia article for 0^0, it gives a bunch of examples for limits of the indeterminate form 0^0, but they all approach different values. For example, lim x to 0+ of (e^(-1/x^2))^x approaches 0, but lim x to 0+ of (e^(-1/x^2))^-x approaches -infinity. The limit lim x to 0+ of (e^(-1/x))^(ax) seems to always approach e^-a, which is not a constant value like 0. So you can't actually find a limit that gives the "correct" value as it approaches 0^0.
This is another example of the definitional difference between something that approaches zero in the limit and zero itself. Various sums that approach zero in the limit will give various values of the limit of "0^0" while strictly 0^0 remains undefined, so there is no "correct" value to it.
On the flip side (taking the inverse) of this is the fact that infinity exists outside the real numbers, so various sums approach infinity in the limit but they do not strictly equal infinity.
I ways trying stuff out and I got the same result as you do. I wanted to find functions 0
Yes, that's why it's called an indeterminate form.
The same is true of others like 1^inf, 0^inf, 0/0, inf/inf etc.
The answers depend on the limit functions you take to get there. This is what defines an indeterminate form.
The purpose of this video is not to show that 0^0 equals anything, but rather that it *can* equal 0 if you set the limiting equations up correctly. I do feel like that should've been made clearer in the video.
Edit: as pointed out below I made a mistake in saying that 0^inf is an indeterminate form
@@alansmithee419 0 to the power of infinity is not indeterminate. However, infinity to the power of 0 is. Also, indeterminate forms yield Aleph-Null as the answer, as we don't know the cardinalities, and also, the answer can be any number in an interval. Indeterminate forms are created because of you are trying to undo an "annihilation" function. An annihilation function yields only one output for all of its inputs, so if an inverse exists, it will have one input but have infinity outputs. However, on any occasion, only one answer can be correct, but because we don't know the cardinalities, all numbers within the interval is vacuously true, as a vacuous truth is defined as if a prerequisite is required to determine the truth or falsity of something, and that prerequisite is not present, we are unsure if it is true, so we will consider it as a vacuously true statement. Therefore, we can consider 0 divided by 0 to be equal to Aleph-Null, with all elements in that set to be vacuously truly equal.
@@alansmithee419 Yes, I was also clarifying that. I think the video was a little misleading, the point is that this a cool limit to solve
I don't follow your channel, and I don't even have to do much math in my everyday job or life. But this legit made me miss calculus for the first time in 15 years. How it felt so much like the art of being clever. This is a beautiful proof.
How can you take the natural log of that limit if it equals to 0? Isn't ln(0) undefined? Isnt that a contradiction in your proof? Or am i missing something here?
Agree, seems that you start with a non-defined operation (Ln(L) is valid only for L > 0, right?). So if your result is L = 0, you start with a contradiction (it seems).
Saw it on desmos from 10^199 to 10^200, the ln(x) function is decreasing but still far from 0 (0.1631), and the square root function is near to 0. Wow! Thanks!
That marker switching is pretty slick
I get the emotional feelings behind this video because learning about some cool math thing that you thought wasn't possible or was really difficult is an emotional experience
Limit form 0^0 is almost always 1. 0^0 is nowadays 1. Very nice example.
Thanks.
Wtf almost always 1? if you take ln both sides and assuming the 0 on the bottom is never negative then lnL=0.ln0=0.-inf=-0.inf, so every 0.inf limit that is not 0 is counter example because e^m /= 1 if m /= 0
@@rafiihsanalfathin9479 that is a theorem. It is almost everywhere 1.
@@rafiihsanalfathin9479idk what you’re saying for a lot of the comment, but what the commenter is saying is that most limits that when plugged in give 0^0 are equivalent to 1. If you take a class that involves L’ Hopital’s rule then you will probably notice this. It doesn’t mean that 0^0 is always equal to one, just that it does for many limits
@@kart338_QK what im saying is that limit that have the form of 0.∞ but have the value other than 0 counter example of what the commenter said. For example lim x->∞ 1/x . -x = -1 (ik this is crappy example but whatever), we can write -x into ln(e^-x) then we got lim x->∞ ln(e^-x)/x=-1 so lim x->∞ (e^-x)^(1/x)=1/e. In general any limit that have the form 0.∞ with the value other than 0 is a counter
14:32 "this right here, it's like the biggest zero.. But! If i have 1/ln of x, that was still not enough. So.. i put another one, and it worked!"
Great idea, brilliant solution, deserved joy of Eureka, deserved like ❤
Thanks!
This is shocking and fascinating, thank you!
not as shocking if you consider this much simpler and more obvious example:
f(x) = e^x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
@@opensocietyenjoyerno, it is shocking this limit form (0^0) is almost always one.
@@lolerie Maybe it is shocking to you...
When (an) is any sequence convergent to 0+, obviously the sequence (an)^(-1/ln(an)) tends to 1/e. It follows that
- if (bn) is a sequence that goes to 0+ significantly faster than -1/(ln(an)), then (an)^(bn) goes to 1,
- if it goes to 0+ significantly slower than -1/(ln(an)), then (an)^(bn) goes to 0.
And obviously the limit can be made equal to anything, it's just a matter of how (bn) compares to (-1/ln(an)).
By assigning L := lim(...), it acquires a fixed value (which you hypothesize to be 0). In that case, taking ln(L) is invalid, because ln is not defined at 0.
On a separate note: have you tried visualizing x^y in 3D space? It might give a visual intuition at least. I'd be curious to see a multi-variable limit calculation of z = x^y, x->0, y->0.
Exactly what I was going to write
This is only a problem in the sense that it highlights the difference between a limit approaching zero and being equal to zero. By setting L :=, he is not saying L is literally ' 'equal to' but that the value of L is assigned the value of the the approachment. Recall, that the definition of a limit doesn't assign a value to the limit. In this case, for all epsilon > 0, there exists a delta > 0 such that ... L < epsilon.
@@rajeevram4681 what? of course limits have values; that's the whole point. otherwise integrals wouldn't have values. the expression on the inside can be said to approach the limit, but the entire point of the lim operator is to evaluate that limit
May I suggest purchasing refillable dry erase markers? Perhaps, if I may be so bold, one black and one red? They write much nicer and more consistently. They are cheaper in the long run for someone who uses whiteboards often. They are better for the environment. The nibs are replaceable as well. I got some that are made by Pilot. They're amazing. Edit: I see you used a blue one in there, so go for it! You earned it with this proof.
I as a student and long time viewer of your videos am very proud. i followed you with many gmails and you really inspires me thank you
i was so angry about the minus sign i almost screamed at you
You must have because I heard you like you were right outside my window!
DAMN!!!!! ❤
Man this was amazing to watch. Idk how u do it but u make math really fun
I prefer the argument for 0^0 being 1. Consider f(x) = x^x. f'(x) = x^x (lnx+1). Roughly:
We examine lim(x→0+) of f(x). We can see that the sign of f' near 0 is < 0:
Let D (delta) be positive. If D is small enough, ln(D) < -1, ie, ln(x) < -1.
So ln(x)+1 < 0.Then it's also true that x^x (ln(x)+1) < 0.
Since f' is negative for small enough D, f(x) is finite increasing as x approaches 0 from the right. And as it does, f(x) gets closer and closer to 1. So f(x) has a definite limit, which, I submit, is 1.
This isn't an agrument its just a limit he found. Limits are never the actual answer to the exact number, so don't worry.
Why are you examining the function x^x, and not x^y? It's not like base and power should always be equal to each other. Sure, if the only case where you use powers satisfies this, then this argument works. But in most cases this restriction is too strong, so you need to look at function of 2 arguments f(x, y) = x^y.
You're right. Approaching 0 in 2 ways is better!@@nbvehbectw5640
Your argument doesn't imply 0^0 *being* 1. It implies *approaching* 1 *if* the function x^x is used
I would love to see a video on this argument!
congratulations! i can see how emotional you were, especially at the end.
The ending was hilarious, i know that feeling 😂
😂
I was about to call you out but you seen your mistake and corrected. Good job dude. Keep it up.
Thank you!!
The usual way to make 0⁰ approach any positive number C (at least the way I usually do it) is to take the limit of (e^(−1/|x|))^(−ln(C)×|x|) as x→0. Maybe this is not a good example in that the expression immediately simplifies to C, so there's no real work in taking the limit, although at least neither the base nor the exponent is constant this way. But of course it doesn't work for C=0.
This is mindblowing, no MINDBREAKING even! Incredible work man!
Finally, I can be watch a daily upload!
Btw I wanna say that you are my hero. Because of you I have found my love for math and am commited to going into theoretical physics. Thank you. ❤
Well done, especially with the stage walk-off at the end (mike drop!). On the one hand, 0^0 can be any non-negative number, so one can say that 0^0 is undefined. On the other hand, 0^0 can be defined to equal 1. This definition makes the most sense, since it removes the discontinuity in functions like x^0.
Limx->inf (sqrt(2x+1)-sqrt(x))=Limx->inf((2x+1-x)/(sqrt(2x+1)+sqrt(x))=Limx->inf((x+1)/(sqrt(2x+1)+sqrt(x))=inf because a linear function grows faster than a sqrt function
Let a=1/2 and x>0. (2x+1)^a-x^a > (2x)^a-x^a = (2^a-1)x^a. Since the right side goes to infinity as x grows, so does the left side.
Also, a simpler example is the limit at zero of f^g where f(x)=e^(-1/x^4) and g(x)=x^2.
I love how you can tell that he is very proud of this :D
They should definitely show something like this to everyone in calc 2, very good example.
Dude…I really like this, well explained and congratulations on figuring this out! ^_^
Thank you!!
it should take you a minute to find a much simpler example:
f(x) = e^-x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
@@opensocietyenjoyer That doesn't work. In order for f(x) to approach 0, you need x approaching negative infinity. However, you can't have x approach negative infinity when talking about x^(-1/2).
my bad, i forgot a minus sign: it should be e^-x
@@MuffinsAPlenty
lim((1/(e^x))^(1/ln x)) also goes to 0 as x approaches infinity. And lim((1/x)^(1/ln(ln x))) do that too. And so on. Just get some very slow decreasing function for exponent and very fast decreasing one for base.
I'm so glad that after 6 years, 0^0 finally decided to overcome his shyness and approach 0. I hope they will live happily ever after.
it was a limit anyways,
so basically the whole crux of the limit was that a smaller number i.e. base(
Having that negative sign return back gave me more relief than actually getting to see a 0⁰ form of limit
Amaze!
Thank you!
So glad this video popped up in my feed!! Great video with explanations (watching your first video actually)!
Hey BPRP =)
Considering how many people noticed the missing minus symbol, you are obviously doing a great job.
Your presentation is well organised so it is easy to spot a mistake and your viewers are well enough educated to understand the error.
Your are awesome and this fake proof looks very convincing :D
I will have my students try to spot the false assumption ^^
Thank you for the nice words! However, I am not sure what you mean by "fake proof". This video isn't about "show 0^0 equals 0", it is about "a limit with the indeterminate form 0^0 being 0". You can also check out my other videos that 0^0->1 and 0^0->e.
Cheers!! : )
@@blackpenredpen The fact that a->0 and b->0 doesn't ensure a^b -> 0^0
Interesting and I did not know that. Do you have an example of this? Thanks.
"this fake proof looks very convincing". Oh my. The math community has some bite.
@@ZaikaNoSeidoikr
Amazing - BlackpenRedpen! And Bluepen! Thank you for this!
10:28 I am just curious... did you drop the negative? Shouldn't it be -xlnx ? Since you multiplied by -1 to cancel.. and the cancel would result in -1 in the right numerator.. ?
EDIT - he fixed it. Thank god.
I'm probably missing a point here, but I think there's a simpler way:
lim x->infinity of ( (x^(-x))^(1/x) )
both x^(-x) and 1/x approach 0 as x approaches infinity
(x^(-x))^(1/x) simplifies to 1/x which is already known to approach 0.
That 2017 video: Can 0^0 approach 0?
th-cam.com/video/Gcl_9KIdpso/w-d-xo.html
Why don't you use epsilon-delta proof to show this limit is 0. But this example is very nice. It's necessary, but not sufficient.
Actually, I take that last sentence back - these are two different functions. And convergence is a property of the function. Even if their behavior seems the same at the point, it does not mean if one converged so will the other.
You'd have to show that somehow there is an upper and lower bound error that converges to zero. Do this proof. Then if checks out, you might have a claim. But by the time you do that, why not go back to epsilon-delta proof.
i have a doubt about the premise of the problem. If x->+inf everything works nice, but 0 as a number can both be reached with a positive limit and with a negative limit. If you plug in -inf in the limits, it doesn't work. I just didn't quite understand this.
@@enderforces7013 With these particular functions (f(x) = √(x+1) - √x , g(x) = 1/ln(ln(x)) ), we can't reach 0 from the negative side.
For x
@@yurenchu still, wouldn't it mean that the limit isn't defined in R?
@@enderforces7013 Which limit? The limit of [f(x)]^g(x) for these particular functions f(x) and g(x) as x goes to +infinity _is_ defined, namely it is 0 . Just as, for example, the limit of (1/2)^x exists for x --> +infinity , even though it doesn't exist for x --> -infinity.
But you may have a point: can we find functions f(x) and g(x) such that the limit of [f(x)]^g(x) is 0 when f(x) and g(x) simultaneously approach 0 ; not only when f(x) and g(x) approach 0 from the positive side but also when f(x) and g(x) approach 0 from the negative side?
blackpenredpen, your job is not yet done!
We can generally define that 0^0 would be equal to 1 or 0, but in my opinion define that is 1 or 0 is not true. As how can 0 transform to 1 or any other number, if is lower? So it's generally can be defined as it's result: 0^0 = Undefined. (You can't turn 0 to itself if you're powering it to a higher number, so that's why is undefined)
Hi, I have a mathematical question. I'd be happy if someone will help me with it. If you use Euler's identity, you can see that e^(iπ) = -1. Now, square both sides to get e^(2iπ) = 1. Now take the natural log on both sides, and 2iπ = 0. And now, divide by 2i to get π = 0. How is this working?
im not even taking calculus yet but my guess is that ln only takes the principal value of it because with imaginary numbers exp function is cos + isin
its like how 0 is not the same as 2pi just because they have the same cos value
ln(e^2ipi) is not the same ln as ln(1) (I THINK. IM NOT AN EXPERT TAKE THIS WITH A GRAIN OF SALT). ln can be treated as the inverse of e^x when dealing with complex and imaginary values and not a simple log function, so you are not performing the same operation to both sides of the equation I don’t think. Again, this is almost certainly inaccurate somewhere considering I’m not a mathematician.
Credit to Akiva Weinberger
"On the complex numbers, the logarithm isn't a function; rather, it's a multifunction (returns multiple values for one argument). This is how e^(2πi)=e^(0)
doesn't imply 2πi=0 after taking logs; ln(1) is all integer multiples of 2πi"
In complex world we dont use just ln, we use Ln (starting from the capital letter). They’re quite similar, but Ln produces infinite amount of outputs for one input
Actually, there are more functions in complex analysis which are analogous to normal ones and they are distinguished by that capital letter
Remember that Euler's formula tells us that e^(iθ) = cosθ+i.sinθ.
So, when we evaluate e^(i2π), we get cos(2π)+i.sin(2π), which gives us 1+0=1.
But we also have e^(i2kπ) = cos(2kπ)+i.sin(2kπ) =1 for k ϵ Z.
Because Cosine and SIne are cyclic with period of 2π, any "inverse" of them will not be a function. Recall that invertible functions must be 1-1 and onto.
So, we can't really have a usual kind of inverse of exponentiation (logarithm) when using complex powers.
The best you can do is recognize that seeking the inverse of a complex exponential will generate an infinite set of solutions of the form a+i.(b+2kπ) for k ϵ Z and a,b ϵ R.
As noted by @user-yy7bq1zx8r, this Complex Logarithm is notated using a capital L (Ln or Log).
Have a look at the Wikipedia article on Complex Logarithm to start digging deeper.
en.wikipedia.org/wiki/Complex_logarithm
From now on, this is my favorite limit
We can only seperate the limit if both limits exist. In this case since lim(lnx) goes to infinity as x goes to infinity, the limit does not exist. So the seperation does not work here. (Or am I missing something?)
Doesn't a limit not exist only when the limit can't converge? Like x->infinity for sin(x)?
I picked up on that but it still approaches 0 seemingly, just need a slightly more rigorous proof.
@@legendgames128 as I know, if a limit does not converge then it is divergent. So still, the limit does not exist.
Yep, this is what I was thinking.
I was searching for this comment lol all the while thinking “am i missing something”
What happens with the "-" sign at 9:30 ?
Oh i should watch the video first😂
Can you use L’hôpitals to conclude anything if the resulting derivative tends to infinity? I don’t believe the proof holds for that case
Okay I'm pretty sure this is incorrect because of one thing.
lim(f(n)) != f(lim(n))
But I might be wrong, this is just an iirc.
A legitimate concern! Here's a video I found that explains it: th-cam.com/video/H2RQC4PxEM0/w-d-xo.html ("When can we switch the limit and function?" by Mu Prime Math).
I'm in the obligation of congratulating you for the massive amount of effort put on this video and solve one of if not the most confusing undeterminations in math.
Amazing work and awesome video!!!🎉🎉🎉
I not sure it is the resolution, but is a solution only for this function.
@@fabiod.674 this function doesn't represent itself but a group of functions like a archetype of functions (you can add a infinite amount of constants in a couple of places and will be limited by 0 anyways) and what it proves is that 0^0 is in fact limited by 0 in some cases (this kind of cases).
I’m confused about a certain step, when you take the natural log of both sides of the equation lim … = L, aren’t you presupposing that L is a number?
it's a minor abuse of notation, but you can do all the same work as e^(ln ...) inside the limit and it comes out exactly the same
This channel is one of those things I do not understand at all, but in a few years will watch again and say, "pretty easy"
Absolute genius. Now show 0^0 can approach i
Shame on u for copying other's things instead of thinking it urself
great video, I had already started typing you forgot the negative!!! but then as I was about to post you noticed it lol
Even with all the effort you took to get that thing to approach 0 it is worth noting that it is an incredible slow function. It actually has a positive slope until it hits around 50, and after that it just goes down glacially slow. The y value is still at 0.005 when x is at 10^15. That is insane for a function that approaches 0, especially when you consider that that hump only peaks at around 0.143.
Thanks!
Thank you!
Lim of sqrt(2x+1)-sqrt(x) is infinity as x goes to infinity
👍
Yup. Because you'll basically have x/sqrtx
Remarkable- g00d explanation of the types of Infinity and corresponding types of zero.
4:58 ln(x) is not defined at x=0, so you can't take the natural logarithm of both sides.
maybe he could try abs val?
The quantities are only approaching 0, not exactly 0. So ln is fine.
@@Ninja20704No the lim operator returns the value it's approaching to, so lim_(x to 0; x > 0) ln(x) is not equal to ln(lim_(x to 0; x> 0) x)
this is just the properties of logarithms and it has nothing to do with the value 0. ln(a^b) = bln(a)
@@jjjannesthe limit does actually go to 0 though, albeit very slowly. Could you set L(x)=sq(x+1)-sq(x)..., then do ln on both sides, then take the limit?
Dude effectively proved that 0^0 is undefined. I bet you can construct limits of 0^0 =
Big grats!
A much simpler example is 1/(x^x) to the power of 1/x. (x goes to infinity as in the video).
Seems legit😂, how did he not see this or are we hallucinating
It's not fun. It all cancels out immediately. The point is to see a limit where it's not immediately obvious that it goes to 0 and yet it does
THIS is so goood! before you wrote the next line i KNEW you’d forget that negative, ive made similar mistakes to that
As a long time viewer since before your channel became so popular, I love to how passionate you were working this out!
I love exploring exponents of zero myself and was in the middle of writing up an idea for working with exponents equal to or less than zero. I stopped working on it after dropping out of a mathematical physics bachelors degree during the height of covid isolation and my poor mental health. After seeing this video I have to ask, would you be interested in talking to me about it?
wow I learned so many math rules in just 15 minutes! the hospital rule is strong. thank you for your demonstration. I hope you can read this, from a French math enthusiast
I have no clue who that guy is and my math isn't good enough to understand everything yet but just listening to him makes me like math even more
That ending with the Mic (pen) drop 😆 Congratulations! I was all giddy when I saw the video title, I knew it was going to be a treat 🤩
it should take you a minute to find a much simpler example:
f(x) = e^x → 0
g(x) = x^(-1/2) → 0
f(x)^g(x) → 0
The confusing here is that we aren't actually evaluating 0⁰.
We are evaluating the limit, L, of the function f(x)^[g(x)], where f(x) and g(x) approach 0, as x approaches 0.
We aren't saying 0⁰ approaches one particular value and based on the choice of f(x) and g(x) the limit L appear to be able to take any value we want it to.
Yes that's what inderminate form means kinda
It's not confusing at all lol. He literally has "limit" on the title and the whiteboard the whole time...
Exactly
I looooooove the way that you were explaining. That was so cool. With that negative, I was just dying 😅. And I totally understood your feelings when you finished it.
Sir i am very weak in maths how i improving in math and start calculas pls sir say something
get better.
get gud
imo calculus is just formulas; make sure you are good with algebra and a little bit of trig (understanding trig identities and understanding unit circle)
Watch my videos. 😃
Study math
This is literally my favorite video on youtube now.
So the problem you faced for so long, and have at last solved, was to find f and g such that the limits as x-> 0⁺ of f(x) and g(x) are both 0, while that of f(x)^g(x) is also 0.
Congratulations!
It seems like there should be a simpler solution, but perhaps there isn't.
Fred
Thank you, Fred. Definitely a satisfying feeling!
@@blackpenredpen Yes, and rightfully so. Meanwhile, I'm trying my hand at other solutions. BTW, I misstated the problem in my comment. Should have said
"... limits as x->∞ ..."
I think they are essentially equivalent, though, by simply replacing the argument of f and g (i.e., x) with its reciprocal, 1/x.
This limit was fascinating! Great job BPRP!
By using calculator to estimate the limit 0^0:
0.00000000001^0.00000000001=0.999999999 (very close to 1)
-0.00000000001^-0.00000000001=-1
As a result, I suspect limit of x^x (x -> 0) doesn't exist
You can't take the limit of x^x as x -> 0^-. (-1/2)^{-1/2} doesn't exist. Neither does x=-1/4, -1/6 and in general -1/2n (and many other values).
I was about to write a comment about the negative sign but I decided to see the result first because it might not be that important. He put it back in and my heart went back to its place.
What is Chain Rule do you mean chen lu?
Though I have studied Calculus for 1 semester in the past, this video still left me scratching my head in confusion as to why it works.
It feels like we missed something. When we use so many tricks it is easy to make a small mistake. My first guess would be that we played too much with infinity for this to work. But I'm not sure... Maybe we could start a discussion about this, or maybe you can make a video going in depth about what could be wrong with this derivation? Thanks for a cool video tho!
Yeah, take a look at 2:18. If you carry out the multiplication on the numerator, this is X+1-X = 1 identically. But not so if you look for the values of the polynomial. For example (1+1)^1/2-(1)^1/2 is not equal to 1. Neither is (1+1)^1/2+(1)^1/2. I think it’s important that the square root of a polynomial is not a polynomial. We could convert the sq roots to a polynomials to actual polynomials with a Taylor series but it might not be necessary. If you plot [(x+1)^1/2 - (x)^1/2] and [(x+1)^1/2 + (x)^1/2 you get two functions that cross, form a node. So you don’t get a single value. It is indeterminate. I think this is explained better with some basic concepts in algebraic geometry, but I would have to review it. But yes, I think this derivation of 0^0 might be more complicated.
@@michaelzumpano7318not sure if I understood your point correctly, but he's just using the special product formula:
(a + b) * (a - b)
= a^2 - b^2
Thus:
(sqrt(x + 1) + sqrt(x)) * (sqrt(x+1) - sqrt(x))
= sqrt(x+1)^2 - sqrt(x)^2
= x + 1 - x
= 1.
It's not that complicated…
Doesnt work at 11:15 he's assuming that lim of infinity/infinity is 1
@@light_asaii4858 I'm sure in some cases you can do that. In most cases actually. Actually, only if the fraction can be simplifed to 1, then you take the limit after, and because there is no 'x' left, the limit become what is left, which is 1 in this case.
@@light_asaii4858No, he’s not assuming that. It’s the result of the limit. You can verify it yourself with wolfram.
I love your enthusiasm man keep up the good work :)
It's not a mic drop moment - he SPIKED it like he scored a touchdown!
(Exits stage left!)😮😮😮😮😮
Stand proud and flex your testicles sir - BRAVO! 👏👏👏👏👏
what
I was about to complain he left out the (-1) from what he factored out, until I watched the video to the end. The lesson of today, be patient to the end before posting a comment.
Over all, awesome video, awesome explanation, some what easy step to follow.😎👍🏽
In set theory, 0^0 = 1, and no analytic limit can change that! 😇
Doesn't matter for any limit when limits only approaches instead of actual there or adding to.
In signals processing, we use 0^0 =1 as well. Otherwise, some important assumptions brake... or that is what I remember, hahaha