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.
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.
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
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."
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).
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
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.
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
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?
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.
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.
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
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
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.
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.
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
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!
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.
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.
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
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.
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!).
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.
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.
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.
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.
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...
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?
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.
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.
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 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
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😋💖💖💖
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
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
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!
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 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.
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.
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.
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.
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.
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)
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
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.
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
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.
Black penned pen, I got moved up to the top class in maths, all thanks to your videos! You are an awesome UA-camr!
That's so awesome!! Thanks for letting me know!! Great job to you as well!!!
@sunnygames4003Black penned pen 🧐🥸
You and 3Brown1Blue are the best UA-camrs for Math. You teach things better than in school. Very good job, greetings from Italy.Peace
Bprp vs 3b1b
Math-off
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.
Kai and Chanelle Simmons you're very welcome, thank you for your nice comment!!
'In your exam do not write "DO MORE WORK", do the work and give the answer' 😂😂
Brings to mind that classic cartoon: "... and then a miracle happens ...".
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
Happy college
0:30
your feelings are irrational
Me : No the answer can't be zero it must be 1.
Bprp: calm down dude we have found lnx to be zero .
What?
0⁰ = 1 doesn't only work as a limit, but as an actual value.
nah, 0^0 = 1 is by a definition, it's similar with 0!=1
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."
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).
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
Kumar No, because it is a limit, so you are never actually multiplying by 0
You haven't invoked the limit yet
While it is true that 0^0 is an indeterminate form, I thought out of "convention", we say 0^0 is 1.
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.
i thought about the 2nd way to solve it! we usually work in class with that form
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
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?
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.
After plugging -0.000001 in Wolfram Alpha, it seems like it converges on -1 (the answer was something like -0.9....+0.00001i)
@@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
I think an honest answer is you can, but at that point we've moved past simple calculus.
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?
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.
Function value is different from limit.
The two need not to be equal.
0^0=1 is the only reasonable definition.
when the blue pen comes out, you know real maths is going down
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
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
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.
0^0 is different from limit of 0^0 form.
Don't be confused.
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.
You say that as if the indeminate form 0^0 could only end up as either 0 or 1. Not true.
@@Cobalt_Spirit root x is getting smaller much faster than x, therefore you have a small number to an even smaller number
ive always thought about sqrt(0) this is a great video thanks!
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+
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
log(a,b)=ln(b)/ln(a)
problem?
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
@@NoNameAtAll2 I know but I just wanted bprp to do it because I know the answer already. I know the identity
Shoot a video about what is t: a ^ b = b ^ a * t
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!
Wolfram showed up with something about (log(2) - q-digamma(1/2, 0, 2))/log(2)
anything but a simple 1/sqrt(e)
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.
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.
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
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.
0^0 is undefined?
ok Let's define it : I put 0^0 = 1 X)
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!).
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.
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.
Could you calculate the same limit, but from the left hand side, lim f(x) where x -> 0- ?
I would “imagine” this is possible.
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.
Black pen red pen please make a video on x^1/lnx as x~0.👍
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.
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.
can we have a 0^0 situation where it's not equal to 1?
Yes
Yes we can : lim(x-->0) 0^x = 0
@@zblxst9347 only right side limit or zero plus. You can't evaluate limit as x goes to zero minus of 0^x
@@kyro1197 oh thanks I didn't see till end
@@IoT_ Obviously we can't. I didn't think about it but it's a good precision
Is 0+ kinda like 0 + dx? If not how does it differ?
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.
How I wished YT was available 40 years ago... LOL!
Thank you
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.
And as always, a great video.
What is (i!) ?
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...
jjeherrera Black Pen Red Pen already addressed this
I have an equations
X^X+X=1
hum 0 ?
Yes but... there is one more answer :)
0? 0^0 is undefined so I would argue 0 is no solution
@@rafciopranks3570 yeah but i think it's an irrational number x=0.3036591268776...
Marcel No, 0^0 is generally taken by mathematicians to be 1.
this is beautiful
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?
*Why people say still say it is undefined*
That is a REALLY REALLY good question that nobody has been able to answer correctly.
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.
Thanks sir..very nicely explained ❤️🤗
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.
Search "L'Hôpital's Rule"
I am proud of myself, I did everything here except setting the limit as L and just putting it as e^(limit)
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
3rd method: do it numerically. Compute 0.0000000000000001^0.00000000000000005
4th method: graph it.
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
what happened at 8:38?you said it will be -1/2x^-3/2 but then you wrote it down -1/2x^3/2
I can't stop smiling. ❤
0^0 is actually equal to one as a computational question. One way to prove this is using blackpenredpen's "BEST FRIEND" formula
Exactly. Using limits cannot disprove this.
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?
tonyotag Not quite. The exponent will be imaginary, but the overall power is complex-valued. You can evaluate it using polar coordinates more easily.
@@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
tonyotag The limit was presented as a problem of real-valued functions, so imaginary numbers are not allowed.
All most all f(x)^g(x) limit -> 1
when both f,g -> 0
Try considering f(x) = |x|^x or f(x) = |x|^sqrt(|x|), then you can take lim x-?0- .
Good but lighting reflected on w borad
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😋💖💖💖
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?
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!
yes
Great video!
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
by definition lim xlogx =0 when x reach 0
I think the second problem can be reduced to lim(x->0)x**x
I love you no humu
Nice
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
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!
Why did you have to differentiate when it was: lim x->0+ (ln x / (1/sqrt x)) ?
L'hopitals rule, look it up
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.
LOL some people just love to show off, don't they.
@@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.
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.
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.
0^(0) = undefined
My school doesn’t allow to use L’hopital
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.
Is there a situation where you have (0+)^(0+) not equalling 1?
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.
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+.
An example where the limit is a (for 00+, and f(x)^g(x)=a->a as x->0+.
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)
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
2³ = 1x2x2x2
2² = 1x2x2
2¹ = 1x2
2º = 1 = 0º
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
No he said limit as x approaches 0 plus which means only the right hand side limit
Me impresiona
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?
No; the exponent is negative because it's in the denominator.
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
According to Wikipedia 0^0 = 1
Both are the same!
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.
Can u plz help me how to integrate e^sinx
But in a video u said 0^0= 1 by convention
In comparison, lim x-> 0+ x^x is 1
Why does my calculator show 1 when I plug in 0^0?????????????
Can I use f(x) = x^x to consider ?????
Yes. I did that before already. That's why I did x^sqrt(x) in this vid.
@@blackpenredpen thanks. Very mush
I see there is an undefined blue pen.
What is the limit for x>0-
The limit DNE because it’s not in the domain due to the square root and the base cannot be negative.
Clever
What’s the minimum of the function?
1/e^2
But why did you not just use the difference quotient, [[f[x] +[h]-f[x]]/h ? I'm lost !
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?
Is it the case that lim(x->0+) x^(x^r)=1 for all r?
With a number that close to 0, and any exponents not equal to 1 or 0, I would round it off to 0
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
0^0=1, don't believe me? use the binomial theorem for (x+y)^n when y=0
0^0 is NOT Undefined!😡
Your Calculus is unusual .