@@PhilosophicalNonsense-wy9gy if it’s the constant 1, then yes. But if it’s some quantity that approaches 1, then it may or may not be. We need to do more work to see.
@@PhilosophicalNonsense-wy9gyyup that's totally correct And I can bet my life on this The real comment in 100% As from left hand side.. you can see (1-)^(♾️)->0 (1+)^{♾️}->♾️ And it's at a state all upto limits
One of the intuitive ways of understanding things like (1+1/x)^x is to think about the "rate" that base --->1 and exponent ---> inf. They are, in a sense, competing against each other. In the end you will be using L'Hopital's rule after switching the ln and the limit.. The ln / lim switch turns an exponentiation to a multiplication: x * ln(1+1/x) Turn that into a ratio moving the 1/x into the denom.. So now you can compare the "rates" (since you now have a ratio) at which both num. and denom. approach 0 by applying L'Hopital to ln(1+1/x) and 1/x. You compare the rates by looking at the ratio f '/g ' - which is ultimately what L'Hopital does. If the ratio of their rates ( f '/g ' ) converges so then will f/g. The inverse of the ln will get you the limit of the original exponentiation. In this case, after L'Hopital, you pick up a factor of (-1/x^2) in both num. and denom. which helps you to simply and then compute the limit of (f ' / g ' ). If (f ' /g ') blows up, then so will f/g.
I think an easy to figure out why the answer is not 1 is if we simply add it together. 1+1/3x = (3x+1)/3x. If we compare the numerator and denominator we can see that as x gets bigger and bigger the numerator is always a little bit bigger than the denominator, which means that the whole expression will not be exactly one as it approaches positive Infinity.
However, why does the limit converge in the first place? My high school teacher told me that (1 + 1/n)^n -> e as n -> +∞ *by definition* , but she never told me why it converges.
e is the base exponent such that derivative equals itself. If you put this constraint in the limit definition of the derivative, you will obtain the limit definition for e.
You can indeed prove that the limit does converge to e however the proof IIRC is long and it is generally not teached in high schools. Typically they will assume that the limit converges to e as a fact and use that to explain the convergence of other limits such as ln(1+x)/x=1 as x goes to 0
1) it monotonically increases 2) has an upper bound therefore, it converges. (1) can be proven if you use Newton's binomial theorem for n and n+1 and compare them (2) can be proven using the above thing + the fact that 1/k!
consider the sequence Eₙ=(1+1/n)ⁿ, and you can prove that it is monotonic increasing, that is Eₙ₊₁≥Eₙ (hint: AM≥GM). Then prove it bounded above by 3. Using MCT, Eₙ converges
Use the formula: lim (1+f(x))^g(x) as f(x) approaches 0 and g(x) approaches infinity = e^lim f(x)g(x) as f(x) approaches 0 and g(x) approaches infinity
i used the exponential property rewriting the function as e to the ln of the function. then brought the x to denomintor as 1/x and applied l hopitals rule once to obtain the cube root of e
I recently studied a formula for this exact situation where 1 is raised to infinity if you try to put in the limit. The format being lim(f(a))^g(x) x->a such that f(a)=1 and g(a)->∞. Then the above limit can be rearranged into the format e^{lim (g(x))[f(x)-1]} x->a
What helps me visuallise the problems is to plot the graph of the limit as they approach large numbers, observing direction will help you avoid basic mistakes!
also you can do this, whenever the there is a exponent and you have a limit like lim f(x)^g(x) = e ^(lim (f(x) -1) g(x)) this would give us e^lim x->infinity(1+1/3x -1)(x) = e^1/3
oh my god I started to do calculus and i was stuck in a problem like this (it was in the eulers numbers limit lol) this video cleared my confusion and i just want to say thanks for the clear explanation
Great video as always. I love to watch your videos. Here, on this limit I would think like this: the "x" inside the parenthesis is bringing the number inside the parenthesis down, BUT the x outside the parenthesis is bringing the final number up. Since one is bringing the number down and the other is bringing the number up, they are fighting against each other, so nothing can be said just by replacing "x" by zero. That is why it's an indeterminate limit.
Another, more general way to do this is to take logs, rewrite it as 0/0 or inf/inf, solve that limit and then take it back to the original. That’s how I did this one. Your way is more elegant and I learned something new!
Would this make more sense if you changed 1 + 1/3x into (3x + 1)/3x ? Cus then if x= 10, then we get (3*10 +1)/3*10 = (31/30)^10 for x=1000 we would get (3*1000+1)/3*1000 = (3001/3000)^1000 for x = 1000000 -> (3000001 / 3000000)^(10^6) etc.
I personally believe that 1/inf=0. It’s something kind of assumed to be true in my field of electrical engineering as well, at least for the most part. The indeterminates require special care though. 0 and infinity are weird. As for 1^inf, that one is interesting. It’s more because x^inf is highly unstable at 1. The limit as x approaches inf from the left is 0, while the limit from the right is infinity. I wouldn’t say 1^inf itself is indeterminate, but (1+0)^inf for sure is. If you can find a case of 1^inf that is indeterminate that isn’t actually (1+0)^inf then I’d really like to see it.
Why is anything e. It's so annoying :D in every math problem there is one step "oh yeah x/y+z^w = e" which is also cos(x)/sqrt(x)*whatyouhadforbreakfast and incidentally it's also everybody's length if intestines divided by the weight of their nose 😅
@@borstenpinsel @williamn3070 You can show that (1+1/t)^t is monotonically increasing in t (using e.g. Bernoulli's inequality) and is bounded above (using e.g. the binomial theorem), so it converges by the monotone sequences theorem. We then define the limit to be e, and you can prove that this is equivalent to any other definition of e you might have had: e.g. to prove it's equivalent to the power series definition, you can again use the binomial theorem and some basic algebraic manipulations (see e.g. courses.maths.ox.ac.uk/pluginfile.php/93465/mod_resource/content/1/supplementary%20notes%20on%20e%20by%20Priestley.pdf).
@@ItamarGlikman It's *one possible* definition of e. Another common definition is the sum of 1/n! for n >= 0, but it's not straightforward to show that these definitions are equivalent.
[ For 1^Inf form, one should learn to convert to a 0 x Inf form by taking ln, and then rewriting in 0/0 form to apply L'Hospital's Rule ] Let the limit be y. Then taking ln (and exchanging lim and ln on RHS), we have ln(y) = lim_x-->Inf x ln (1 + 1/3x) = lim_x-->Inf (ln(1 + 1/3x))/(1/x) This is now 0/0 form, so let us apply L'Hospital's Rule. We will get ln(y) = lim_x-->Inf (1/3 (-1/x^2))/ ((1 + 1/3x)(-1/x^2)) = lim_x-->Inf (1/3) / (1 + 1/3x) and here now we can put x = Inf to get ln(y) = 1/3, which implies y = e^(1/3).
Right? Infinity IS tricky! Which makes it fascinating. It should't be intuitive thinking about something that is infinite, never ending, without bound. Something like that just *doesn't exist* in the real world (AFAIK). That's what makes it beautiful that it exists in math.
@@Thorcoal I think it’s more useful to think of infinity as a mathematical tool than as some nebulous possibly real thing. For example, the relativistic energy of something with mass going at the speed of light is infinity. What does that mean? Only massless objects like photons can travel at the speed of light, anything with mass must be slower
Great video Had to rejog my brain on this problem, it's deceptively hard but as soon as you wrote the e limit (which i already knew and thought of but seeing it is different) everything clicked
When you look at these questions it's clear that as one term goes towards zero, it's contribution to the overall answer also increases. So it's obvious why you can't just treat the term on its own as zero -- because its contribution is modified later by something that increases the value at a relatively corresponding rate. If it were just 1/3x as x->0 then yeah, it's zero.
I think you deserve some recognition for "The Fact". As it is unnamed, perhaps the " BPRP Lemma " would be appropriate. Your contributions to the mathematical world compare to Rick Beato's contributions to the musical world. Thank you Professor.
I would take a simpler approach. If you look at a binomial expansion of (1+x)^n, then the first term is 1, and the second term is n*x. Using that here, the first term is 1, the second term is x/3x = 1/3. That means that overall, the result will be greater than 1+1/3, so it can't be equal to 1.
That is brilliant. I did the problem before watching the video and I got 1 as an answer...of course, I was wrong. Thank you so much for doing this interesting problem.
So the thing is: you can replace 1/3x with 0. Just need to remember that all 0s and all infinities in the limits are not numbers. They're variables. Infinity has a really, really big number as its value, and 0 has really, really small number as its value. And they're written like that just for simplicity. So while you can do something like: 1 + 1/3x = 1 + 0. You can't do 1 + 0 = 1.
because the first time it approaches 1 whereas the second time it is exactly one. You cannot say for certain whether a number approaching 1 to the infinity power is equal to 1
Its because 1, in this case, is actually 1 plus 1/x where x approaches infinity As long as the rest of the problem is finite, it is effectively equal to 1(but not exactly equal to 1, its essentially equal to 1.0000...0001 where there is an uncountable infinity of 0's between the "...". No matter how large the number you multiply (1+1/infinity) by is, you will still treat it as 1x, until the number you are multiplying by is also approaching infinity.
@@EdwardCurrent 1^inf is indeterminate, always. However, we evaluate indeterminate forms all the time, within the context of a given limit of a function. When he says that "exactly 1"^inf is 1, what he means is that the limit of the function 1^x as x->inf is 1. This is the evaluation of a specific limit of a specific function, and so it doesn't necessarily apply to other limits that have the 1^inf form. It's the same as if you have the function x/x and take the limit as x->inf. This is an inf/inf situation, which is indeterminate. However, we can use several methods to show that the value of this limit is 1. So in this situation inf/inf =1, however, for other limits of other functions that end up in an inf/inf situation the solution may be different. That's essentially what indeterminate means: that there is not a single generalized solution, and that other methods must be used in order to evaluate them in the context of a given function/limit. So in general 1^inf is indeterminate. But in the specific context of 1^x as x->inf then it is equal to 1.
6:45 Is it actually 'legal' to shift the cube root outside of the limit? That was raher slight of hand. The very thing that you are trying to expose as being wrong.
Perfectly legal, cube root of x is continuous on every point of the line, in particular on R+, so yes. It could even be a square root, no problem in that specific limit.
I do understand most of it, but it does still bother me that we 1/inf doesn’t simplify to zero. Normally I’d let it go, but I feel like there were situations where that was okay to do. And then to top it off “exactly 1^x” is okay to work with? When x approaches infinity? Hmm. I will have to go back to the drawing board for this one! But that you for bringing this to my attention again! Edit: absolutely terrible phrasing. And thank you to everyone replying! I understand now. Beautiful explanations.
Thing is 0 (as well as ∞) in that case is not a number. It's a variable like a t in the video. It just holds really smal number, which is almost an actual 0, but never exactly it. So while you can do 1/∞ = 0, you can't do 1+0 = 1, cause you're not adding 2 plain numbers, you're adding a number and a variable. 1 + 0 is only approximately 1, which is fine, if there are no further operations that could change this 0 into something other, then just a really small number, but with a (1+0)^∞ you can't do that, cause ^∞ does change it. It's fine because the result of limits is often an approximate value which the actual result approaches, but never equal to. As for 1^∞ where 1 is exactly 1, yes, it is 1, because you can multiply 1 by 1 how many times you wan't, it will not change the answer.
The general replacement of 1/inf = 0 as well as 1/0 = infinity that is somewhat common in calculus, is that these are shorthands that assume that this is the only relevant behavior of the function. But in this case that's not true. We have 1/inf that is pushing the function towards 1, but there's also the exponent that is pushing the function towards infinity. Because these two effects are competing with eachother we cannot adequately describe the overall behavior of the function with this simple replacement. We technically *can* make the substitution, but as noted in the video we end up with 1^inf which is indeterminate and so doesn't help us to evaluate the limit here. The substitution isn't "wrong" it's just not useful. As far as the "exactly 1"^inf part, this is also a bit of a shorthand that doesn't fully explain the reasoning. 1^inf is indeterminate, in all situations. However, as is often the case in calculus, we can still evaluate indeterminate forms within the context of the function/limit we are working with. And as it happens, when the function you are working with is simply 1^x as x->infinity, this limit does evaluate to 1. The way he phrases it with the "exactly 1" wording is a little misleading (though not wrong) in that it sort of hides the actual reason. The actual reason is because it's the evaluation of a specific limit that gives us 1, whereas other functions (such as the one in this video) are different functions where 1^inf may evaluate to a different value. It's the same as how x/x as x->inf is an inf/inf situation that evaluates to 1 in the limit, but other inf/inf limits for other functions may end up having different values.
It would simplify to zero if that was the last step in calculating the limit. But since it's not in this case, but rather an intermediate step in calculating the limit, you have to take into account that the later calculations that will be performed on this intermediate step will so-to-say "cancel out" the effect.
From my perspective, the crux of the issue is that the x in the denominator and in the exponent are the *same* variable, which can interfere with intuition. Two important things to remember: (1) the limit to infinity is defined by what happens to the estimates as we plug in larger and larger values; we can never actually plug in infinity, and even if we could, we actually don’t care what it would be! We only care about the behavior of the values as we get “close” to infinity. (2) We *cannot* increase one x before or without increasing the other-they are the same x. #1 is a definition, so that’s not changeable, but what if we break #2 by forcibly separating them? It actually gets pretty fun: Getting rid of the 3 for clarity, let’s say we have f(x,y) = (1 + 1/x)^y … Then the limit would actually depend on *how* we send x and y to infinity-graphically, you can think of this as following some path in 2D out into quadrant 1, and we can pick the path freely so long as it goes on forever in that quadrant! If we follow the limit of (x,y) -> (inf,inf) along the line y=x, then we get the classic limit of e. If we follow it along y=x/3, we get e^(1/3) like in the video, and in general, for y=mx+b, I’m pretty sure we get e^m. In the special case of evaluating strictly x first then y, or vice versa, the graphical model gets disconnected from the limit a bit, but you can imagine it as following along the x or y axis to infinity in that direction, and then after we “reach” infinity along that axis, we then turn 90° and going in the the other axis’ direction toward infinity. Along x axis: This is doing x -> inf first, then y -> inf, which gets you 1^y -> 1 for the limit. (You can get this even by just plugging in 0 for m in e^m above.) Along the y axis: This is doing y -> inf first, which… uhhh, I think diverges to + or - infinity? (Either way, this is an edge case.)
Let the heat losses of one machine be E=E(1-n) and let each successive machine capture the energy losses from the previous machine, so that the total losses are E=E(1-n).(1-n).....=E(1-n)^x. If the efficiency of the machines is n=1/x and x approaches infinity, then the efficiency of one machine will be zero, but the efficiency of the whole system is not zero E(losses)=E(1-1/x)^x=E.e^-1 and the efficiency E(efficiency)=E(1-e^-1). Similarly, we could consider the efficiency of electric field lines acting on gravitational field lines, where the efficiency is zero, but if their number (intensity ) approaches infinity , then it may not be zero.
Whenever I see infinity expressed mathematically I always treat it as a sorta variable myself, because it can be any positive number really....infinity just means a really really really big number after all.
I'm a dumb dumb welder with a caveman brain, and no practical application for this level of math. That being said, I enjoy your videos and your teaching style. Plus that two marker technque is pretty slick!
Rewrite as: ∫ x^(-5) + 1 dx Integral of a sum is the sum of the integral so: ∫ x^(-5) dx + ∫ 1 dx The second integral is trivial: ∫ x^(-5) dx + x + c The remaining integral can be solved using the power rule: (-1/4) x^(-4) + x + c
@@ronaldjensen2948The problem, as written in the original comment is ambiguous, and can be interpreted as F(x) = (1/x^5) + 1 or as F(x) = 1/(x^5 +1). x^-5 +1 = 1/x^5 + 1 = 1 + (x^5/x^5) = (x^5+1)/x^5. Therefore, they are not equivalent. Considering the second interpretation, the first step in solving this integral would be using the algebraic identity : A^5 + B^5 = (A + B) (A^4 - A^3B + A^2B^2 - AB^3 + B^4), and then proceeding into a long chain of partial fractions.
Sir, you said that x = (1/3)t, but when substituting t in the limit,i.e., in x -----> infinity you should have been wrote it like (1/3)t -----> intinity. Why sir you wrote it as t -----> infinity
If you look at the equation after he subbed t for x, yes. But he did it for the original form of the equation. Remember t and x are just labels, they can be anything and don't actually matter -- they're only different so that we can tell them apart.
No. lim_t→∞(1 + 1/t)^t is not e BY DEFINITION, but by result. Don't they teach the binomial expansion any more? (1+ a)^n = 1 + na + n(n-1)a^2/2! + n(n-1)(n-2)a^3/3! + ... Set n = 1/a = t and you get: (1 + 1/t)^t = 1 + 1 + t(t-1)/t^2/2! + t(t-1)(t-2)/t^3/3! + .. As t→∞, we can see that t ≈ (t-1) ≈ (t-2), etc. Therefore t(t-1)/t^2 ≈ 1, and t(t-1)(t-2)/t^3 ≈ 1, etc. So we get: lim_t→∞(1 + 1/t)^t = 1 + 1 + 1/2! + 1/3! +1/4! which is e _by definition._
There are two main definitions of e which are equivalent (and that's a theorem you can prove), one with the power series and one as the limit of the *sequence* a_n = (1+1/n)^n as n->infty. It's important to notice it's the limit of a sequence and not of the function (1+1/x)^x as exponentiation with real exponents is only defined through the exp function (basically after defining e and proving some properties)
Also, beware of "substituting" t(t-1)...(t-k+1)/t^k with 1 as, while it would be legal in a limit of a sequence by the theorem of substitution (not sure about the name, I learnt about it in another language) here you're considering an infinite sum, and as you probably know the infinite sum of infinitesimal errors can be non infinitesimal (well, in this case it still works, but you need to put more work to prove it)
@@notmymain2256 The most distinctive definition in my opinion is the base of logarithms whose anti-derivative is 1/x. But of course, there are lots of "definitions" that can be derived from each other. Nevertheless, it is rarely helpful to start from the limit as you suggest, since it is trivially found equivalent to the power series by use of the binomial theorem. The limit is, of course, the same whether it is approached through the integers or the reals. Your point about exponentiation with reals is another reason why using limits is a less satisfactory starting point for a definition.
have I passed calculus several years ago? yes. have I already knew the answer? yes. have I watched the video till the end because of the brilliant and easy explanation? also yes.
Practice and memory. Then you recognize patterns. "This looks like the formula for e, can I use that?". Some math youtube channels skip this part, and just say "let us write this a different way for no reason". That is not how it works. This channel is much better than most in explaining WHY you do those first interesting and sometimes creative steps. The rest is following the rules and seeing where you end up.
Rather than explain it in terms of indeterminate forms, I would say that lim f(x)^g(x) = (lim f(x))^(lim g(x)) only works if lim f(x) and lim g(x) both converge. In this case lim g(x) doesn't converge.
306 / 5 000 hello, I love what you do however I would like you to add something more to your mathematics presentations such as the possible and applicable use of your equation in a specific field like chemistry, measurements, probability or related mathematics to our daily life, have a good day
Get your indeterminate cat t-shirt: 👉 amzn.to/3qBeuw6
DO NOT BE CONFUSED!
The number 1, to the power of infinity, is 1.
A number that approaches 1, to the infinity, can be anything.
1^∞ = 1?????????
@@PhilosophicalNonsense-wy9gy It's always 1 if you mutliply 1 by itself as many times as you want.
@@PhilosophicalNonsense-wy9gy if it’s the constant 1, then yes. But if it’s some quantity that approaches 1, then it may or may not be. We need to do more work to see.
@@PhilosophicalNonsense-wy9gyyup that's totally correct
And I can bet my life on this
The real comment in 100%
As from left hand side.. you can see (1-)^(♾️)->0
(1+)^{♾️}->♾️
And it's at a state all upto limits
@@PhilosophicalNonsense-wy9gy yes, please see 3:40
One of the intuitive ways of understanding things like (1+1/x)^x is to think about the "rate" that base --->1 and exponent ---> inf. They are, in a sense, competing against each other. In the end you will be using L'Hopital's rule after switching the ln and the limit.. The ln / lim switch turns an exponentiation to a multiplication: x * ln(1+1/x) Turn that into a ratio moving the 1/x into the denom.. So now you can compare the "rates" (since you now have a ratio) at which both num. and denom. approach 0 by applying L'Hopital to ln(1+1/x) and 1/x.
You compare the rates by looking at the ratio f '/g ' - which is ultimately what L'Hopital does. If the ratio of their rates ( f '/g ' ) converges so then will f/g. The inverse of the ln will get you the limit of the original exponentiation. In this case, after L'Hopital, you pick up a factor of (-1/x^2) in both num. and denom. which helps you to simply and then compute the limit of (f ' / g ' ).
If (f ' /g ') blows up, then so will f/g.
that second paragraph suddenly made l'hopital make sense to me, i can see the justification behind it now
I think an easy to figure out why the answer is not 1 is if we simply add it together. 1+1/3x = (3x+1)/3x. If we compare the numerator and denominator we can see that as x gets bigger and bigger the numerator is always a little bit bigger than the denominator, which means that the whole expression will not be exactly one as it approaches positive Infinity.
That is some quick marker switching at 6:03!!
However, why does the limit converge in the first place? My high school teacher told me that (1 + 1/n)^n -> e as n -> +∞ *by definition* , but she never told me why it converges.
e is the base exponent such that derivative equals itself. If you put this constraint in the limit definition of the derivative, you will obtain the limit definition for e.
You can indeed prove that the limit does converge to e however the proof IIRC is long and it is generally not teached in high schools.
Typically they will assume that the limit converges to e as a fact and use that to explain the convergence of other limits such as ln(1+x)/x=1 as x goes to 0
1) it monotonically increases
2) has an upper bound
therefore, it converges.
(1) can be proven if you use Newton's binomial theorem for n and n+1 and compare them
(2) can be proven using the above thing + the fact that 1/k!
consider the sequence Eₙ=(1+1/n)ⁿ, and you can prove that it is monotonic increasing, that is Eₙ₊₁≥Eₙ (hint: AM≥GM).
Then prove it bounded above by 3.
Using MCT, Eₙ converges
you can also observe (not prove) this by substituting large numbers into n with your calculator to see that it converges to e
Use the formula: lim (1+f(x))^g(x) as f(x) approaches 0 and g(x) approaches infinity = e^lim f(x)g(x) as f(x) approaches 0 and g(x) approaches infinity
i used the exponential property rewriting the function as e to the ln of the function. then brought the x to denomintor as 1/x and applied l hopitals rule once to obtain the cube root of e
I recently studied a formula for this exact situation where 1 is raised to infinity if you try to put in the limit. The format being lim(f(a))^g(x)
x->a such that f(a)=1 and g(a)->∞. Then the above limit can be rearranged into the format e^{lim (g(x))[f(x)-1]}
x->a
Do you have the demo for it please?
What helps me visuallise the problems is to plot the graph of the limit as they approach large numbers, observing direction will help you avoid basic mistakes!
also you can do this, whenever the there is a exponent and you have a limit like lim f(x)^g(x) = e ^(lim (f(x) -1) g(x))
this would give us e^lim x->infinity(1+1/3x -1)(x) = e^1/3
Absolutely this is the easiest way to solve this question in literally 2 steps
do you have the demo for this formula please?
@@voidete7793 i have already written the formula in thr comment
For the demo, you can try using it in the same question as in the video
Another way to think about it: 1.000000000000000001 raised to x goes to infinity as x goes to infinity as well.
The base is slightly larger than 1. Consider something like (1+0.1)^10, it's not 1, there's a rate of interest 10% compounded over 10 years :)
Nice link!
oh my god
I started to do calculus and i was stuck in a problem like this
(it was in the eulers numbers limit lol)
this video cleared my confusion and i just want to say thanks for the clear explanation
When I begin calculus, I am certain I will be watching a lot of your videos. Good content!
Great video as always. I love to watch your videos. Here, on this limit I would think like this: the "x" inside the parenthesis is bringing the number inside the parenthesis down, BUT the x outside the parenthesis is bringing the final number up. Since one is bringing the number down and the other is bringing the number up, they are fighting against each other, so nothing can be said just by replacing "x" by zero. That is why it's an indeterminate limit.
That’s surprisingly interesting, never thought about it
Another, more general way to do this is to take logs, rewrite it as 0/0 or inf/inf, solve that limit and then take it back to the original. That’s how I did this one.
Your way is more elegant and I learned something new!
Beauty of algebraic manipulation. I had a feeling it was related to e.
Would this make more sense if you changed 1 + 1/3x into (3x + 1)/3x ?
Cus then if x= 10, then we get (3*10 +1)/3*10 = (31/30)^10
for x=1000 we would get (3*1000+1)/3*1000 = (3001/3000)^1000
for x = 1000000 -> (3000001 / 3000000)^(10^6)
etc.
(2^(1/x))^x is a 1^infinity type limit as x goes to infinity. But clearly the limit is not 1, it's 2.
MATH. IS. BEAUTIFUL.
Can also take ln of each side and use the rule that lim ln = ln lim
I personally believe that 1/inf=0. It’s something kind of assumed to be true in my field of electrical engineering as well, at least for the most part. The indeterminates require special care though. 0 and infinity are weird. As for 1^inf, that one is interesting. It’s more because x^inf is highly unstable at 1. The limit as x approaches inf from the left is 0, while the limit from the right is infinity. I wouldn’t say 1^inf itself is indeterminate, but (1+0)^inf for sure is. If you can find a case of 1^inf that is indeterminate that isn’t actually (1+0)^inf then I’d really like to see it.
This is true but a general formula exists for 1^infinity problems
Lim x->a ((f(x))^g(x))=e^(Lim x->a (f(x)-1)*g(x))
do you have the demo for it please?
why does lim of (1+ 1/t)^t, t--> infinity = e?
Why is anything e. It's so annoying :D in every math problem there is one step "oh yeah x/y+z^w = e" which is also cos(x)/sqrt(x)*whatyouhadforbreakfast and incidentally it's also everybody's length if intestines divided by the weight of their nose 😅
@@borstenpinsel @williamn3070 You can show that (1+1/t)^t is monotonically increasing in t (using e.g. Bernoulli's inequality) and is bounded above (using e.g. the binomial theorem), so it converges by the monotone sequences theorem. We then define the limit to be e, and you can prove that this is equivalent to any other definition of e you might have had: e.g. to prove it's equivalent to the power series definition, you can again use the binomial theorem and some basic algebraic manipulations (see e.g. courses.maths.ox.ac.uk/pluginfile.php/93465/mod_resource/content/1/supplementary%20notes%20on%20e%20by%20Priestley.pdf).
@@borstenpinselbro let me have a sip of your drink
Thats literally the definition of e...
@@ItamarGlikman It's *one possible* definition of e. Another common definition is the sum of 1/n! for n >= 0, but it's not straightforward to show that these definitions are equivalent.
When you become a professional in any field, you see things immediately that those learning might take hours to see, or not see at all.
[ For 1^Inf form, one should learn to convert to a 0 x Inf form by taking ln, and then rewriting in 0/0 form to apply L'Hospital's Rule ] Let the limit be y. Then taking ln (and exchanging lim and ln on RHS), we have ln(y) = lim_x-->Inf x ln (1 + 1/3x) = lim_x-->Inf (ln(1 + 1/3x))/(1/x) This is now 0/0 form, so let us apply L'Hospital's Rule. We will get ln(y) = lim_x-->Inf (1/3 (-1/x^2))/ ((1 + 1/3x)(-1/x^2)) = lim_x-->Inf (1/3) / (1 + 1/3x) and here now we can put x = Inf to get ln(y) = 1/3, which implies y = e^(1/3).
Can we use the logarithmic differentiation rule?
After that we use L'hôpital rule ?
God I hate infinity it’s just so tricky
I love infinity, it’s just so tricky
Right? Infinity IS tricky! Which makes it fascinating. It should't be intuitive thinking about something that is infinite, never ending, without bound. Something like that just *doesn't exist* in the real world (AFAIK). That's what makes it beautiful that it exists in math.
@@Thorcoal I think it’s more useful to think of infinity as a mathematical tool than as some nebulous possibly real thing. For example, the relativistic energy of something with mass going at the speed of light is infinity. What does that mean? Only massless objects like photons can travel at the speed of light, anything with mass must be slower
love the reddit question videos
Great video
Had to rejog my brain on this problem, it's deceptively hard but as soon as you wrote the e limit (which i already knew and thought of but seeing it is different) everything clicked
just go back to the ε-δ definition then everything are clear
So lim->inf (1+1/3x)^x does not have a defined value so you cant use aritmethic for limits and need to try an alternative path
How can we take the limit into the parentheses with cube root
This guy is a ninja when it comes to hot swapping markers on the go!
When you look at these questions it's clear that as one term goes towards zero, it's contribution to the overall answer also increases. So it's obvious why you can't just treat the term on its own as zero -- because its contribution is modified later by something that increases the value at a relatively corresponding rate. If it were just 1/3x as x->0 then yeah, it's zero.
I think you deserve some recognition for "The Fact". As it is unnamed, perhaps the " BPRP Lemma " would be appropriate. Your contributions to the mathematical world compare to Rick Beato's contributions to the musical world. Thank you Professor.
I would take a simpler approach.
If you look at a binomial expansion of (1+x)^n, then the first term is 1, and the second term is n*x.
Using that here, the first term is 1, the second term is x/3x = 1/3. That means that overall, the result will be greater than 1+1/3, so it can't be equal to 1.
That is brilliant. I did the problem before watching the video and I got 1 as an answer...of course, I was wrong. Thank you so much for doing this interesting problem.
So the thing is: you can replace 1/3x with 0.
Just need to remember that all 0s and all infinities in the limits are not numbers. They're variables. Infinity has a really, really big number as its value, and 0 has really, really small number as its value. And they're written like that just for simplicity.
So while you can do something like: 1 + 1/3x = 1 + 0. You can't do 1 + 0 = 1.
5:24 or is it??....🤨
I remember when y saw that limit. I was struggling to resolve it and symbols share me a property that I never seen before
1 + 1/3000= 1.003 1.0003 ; however does not affect video in question.
UA-cam keeps recommending you to me. And tbh as someone who loves maths I finally gotta subscribe to you.
Just Log it out.
Yep. The limit of (1+1/x)^x as x->infinity is e≈2.718, so the limit of (1+1/3x)^x is e^(1/3)
Yes I was thinking about that too
is there a version of 'the fact' for the lim n -> inf (1-1/n)^n = 1/e bc i feel like that might pop up once or twice
Yes. That’s when a=-1 and b=1
had a similar limit but never understood why…. Now I get it thx!!!
Confused at 3:55 -- the shirt says that 1 ^∞ is an indeterminate form, so how does it = 1?
because the first time it approaches 1 whereas the second time it is exactly one. You cannot say for certain whether a number approaching 1 to the infinity power is equal to 1
Its because 1, in this case, is actually 1 plus 1/x where x approaches infinity
As long as the rest of the problem is finite, it is effectively equal to 1(but not exactly equal to 1, its essentially equal to 1.0000...0001 where there is an uncountable infinity of 0's between the "...". No matter how large the number you multiply (1+1/infinity) by is, you will still treat it as 1x, until the number you are multiplying by is also approaching infinity.
@@benseb2512 So the mention on the shirt is specific to the limit, and strictly speaking 1^∞ is not an indeterminate form?
@@EdwardCurrent 1^inf is indeterminate, always.
However, we evaluate indeterminate forms all the time, within the context of a given limit of a function. When he says that "exactly 1"^inf is 1, what he means is that the limit of the function 1^x as x->inf is 1. This is the evaluation of a specific limit of a specific function, and so it doesn't necessarily apply to other limits that have the 1^inf form.
It's the same as if you have the function x/x and take the limit as x->inf. This is an inf/inf situation, which is indeterminate. However, we can use several methods to show that the value of this limit is 1. So in this situation inf/inf =1, however, for other limits of other functions that end up in an inf/inf situation the solution may be different.
That's essentially what indeterminate means: that there is not a single generalized solution, and that other methods must be used in order to evaluate them in the context of a given function/limit. So in general 1^inf is indeterminate. But in the specific context of 1^x as x->inf then it is equal to 1.
@@phiefer3 Very well explained, thank you
1/∞ does not equal 0. It equals 0+. We don't know what it equals. And 1+0+ is just 1+. 1+ ^ ∞ = ∞
Low key tempted to get some of that product drop on EXPO markers. 😅
😆
You can’t assume the it’s exactly zero because than the limit would be 1^inifinity = 1. However that’s not correct
6:45 Is it actually 'legal' to shift the cube root outside of the limit? That was raher slight of hand. The very thing that you are trying to expose as being wrong.
Perfectly legal, cube root of x is continuous on every point of the line, in particular on R+, so yes. It could even be a square root, no problem in that specific limit.
@@Rafaeu777 True, but is this limit finite? If not then all bets are off.
@@alphalunamare Not really, you can use this same argument along with the extension of Infinite limits and limits at infinity
I got the same answer by taking ln of the limit then converting it to a ratio and applying La Hospitals rule.
My intuition says it goes to infinity as it is 1. smthn to the power of inf.
Ha, here I really thought it approaches 1.
But the ever increasing exponent is not to be underestimated.
Actually, a^x is not continous at infinity as neighborhoods of infinity actually grow larger on the riemann sphere, lol...
Can you just multiply the exponent by 3/3 and get the same result? I dont remember limits very well.
why isn't infinity/3 = infinity though?
It is tho
Infinity divided by anything non-infinite or non-zero is still infinity
e^1/3
I do understand most of it, but it does still bother me that we 1/inf doesn’t simplify to zero. Normally I’d let it go, but I feel like there were situations where that was okay to do.
And then to top it off “exactly 1^x” is okay to work with? When x approaches infinity? Hmm. I will have to go back to the drawing board for this one! But that you for bringing this to my attention again!
Edit: absolutely terrible phrasing. And thank you to everyone replying! I understand now. Beautiful explanations.
Thing is 0 (as well as ∞) in that case is not a number. It's a variable like a t in the video. It just holds really smal number, which is almost an actual 0, but never exactly it.
So while you can do 1/∞ = 0, you can't do 1+0 = 1, cause you're not adding 2 plain numbers, you're adding a number and a variable.
1 + 0 is only approximately 1, which is fine, if there are no further operations that could change this 0 into something other, then just a really small number, but with a (1+0)^∞ you can't do that, cause ^∞ does change it.
It's fine because the result of limits is often an approximate value which the actual result approaches, but never equal to.
As for 1^∞ where 1 is exactly 1, yes, it is 1, because you can multiply 1 by 1 how many times you wan't, it will not change the answer.
The general replacement of 1/inf = 0 as well as 1/0 = infinity that is somewhat common in calculus, is that these are shorthands that assume that this is the only relevant behavior of the function. But in this case that's not true. We have 1/inf that is pushing the function towards 1, but there's also the exponent that is pushing the function towards infinity. Because these two effects are competing with eachother we cannot adequately describe the overall behavior of the function with this simple replacement. We technically *can* make the substitution, but as noted in the video we end up with 1^inf which is indeterminate and so doesn't help us to evaluate the limit here. The substitution isn't "wrong" it's just not useful.
As far as the "exactly 1"^inf part, this is also a bit of a shorthand that doesn't fully explain the reasoning. 1^inf is indeterminate, in all situations. However, as is often the case in calculus, we can still evaluate indeterminate forms within the context of the function/limit we are working with. And as it happens, when the function you are working with is simply 1^x as x->infinity, this limit does evaluate to 1. The way he phrases it with the "exactly 1" wording is a little misleading (though not wrong) in that it sort of hides the actual reason. The actual reason is because it's the evaluation of a specific limit that gives us 1, whereas other functions (such as the one in this video) are different functions where 1^inf may evaluate to a different value.
It's the same as how x/x as x->inf is an inf/inf situation that evaluates to 1 in the limit, but other inf/inf limits for other functions may end up having different values.
It would simplify to zero if that was the last step in calculating the limit. But since it's not in this case, but rather an intermediate step in calculating the limit, you have to take into account that the later calculations that will be performed on this intermediate step will so-to-say "cancel out" the effect.
From my perspective, the crux of the issue is that the x in the denominator and in the exponent are the *same* variable, which can interfere with intuition. Two important things to remember: (1) the limit to infinity is defined by what happens to the estimates as we plug in larger and larger values; we can never actually plug in infinity, and even if we could, we actually don’t care what it would be! We only care about the behavior of the values as we get “close” to infinity. (2) We *cannot* increase one x before or without increasing the other-they are the same x.
#1 is a definition, so that’s not changeable, but what if we break #2 by forcibly separating them? It actually gets pretty fun: Getting rid of the 3 for clarity, let’s say we have f(x,y) = (1 + 1/x)^y … Then the limit would actually depend on *how* we send x and y to infinity-graphically, you can think of this as following some path in 2D out into quadrant 1, and we can pick the path freely so long as it goes on forever in that quadrant!
If we follow the limit of (x,y) -> (inf,inf) along the line y=x, then we get the classic limit of e. If we follow it along y=x/3, we get e^(1/3) like in the video, and in general, for y=mx+b, I’m pretty sure we get e^m.
In the special case of evaluating strictly x first then y, or vice versa, the graphical model gets disconnected from the limit a bit, but you can imagine it as following along the x or y axis to infinity in that direction, and then after we “reach” infinity along that axis, we then turn 90° and going in the the other axis’ direction toward infinity.
Along x axis: This is doing x -> inf first, then y -> inf, which gets you 1^y -> 1 for the limit. (You can get this even by just plugging in 0 for m in e^m above.)
Along the y axis: This is doing y -> inf first, which… uhhh, I think diverges to + or - infinity? (Either way, this is an edge case.)
Thank you so much everyone for helping me Understand this!
Let the heat losses of one machine be E=E(1-n) and let each successive machine capture the energy losses from the previous machine, so that the total losses are E=E(1-n).(1-n).....=E(1-n)^x. If the efficiency of the machines is n=1/x and x approaches infinity, then the efficiency of one machine will be zero, but the efficiency of the whole system is not zero E(losses)=E(1-1/x)^x=E.e^-1 and the efficiency E(efficiency)=E(1-e^-1). Similarly, we could consider the efficiency of electric field lines acting on gravitational field lines, where the efficiency is zero, but if their number (intensity ) approaches infinity , then it may not be zero.
Otherwise, a similar concept can be seen in e.g. the equations of thermal diffusion, the Yukawa potential...etc.
Is it just me or were there police sirens in the background
I was taught that when lim a^b = 1^inf then lim a^b = e^(lim b(a-1))
Whenever I see infinity expressed mathematically I always treat it as a sorta variable myself, because it can be any positive number really....infinity just means a really really really big number after all.
No, it is not a number. That is why we have a whole chapter on limits! “Really big number” implies it is finite.
I'm a dumb dumb welder with a caveman brain, and no practical application for this level of math. That being said, I enjoy your videos and your teaching style. Plus that two marker technque is pretty slick!
Please try to integrate 1/x^5+1
Rewrite as:
∫ x^(-5) + 1 dx
Integral of a sum is the sum of the integral so:
∫ x^(-5) dx + ∫ 1 dx
The second integral is trivial:
∫ x^(-5) dx + x + c
The remaining integral can be solved using the power rule:
(-1/4) x^(-4) + x + c
@@ronaldjensen2948The problem, as written in the original comment is ambiguous, and can be interpreted as F(x) = (1/x^5) + 1 or as F(x) = 1/(x^5 +1).
x^-5 +1 = 1/x^5 + 1 = 1 + (x^5/x^5) = (x^5+1)/x^5. Therefore, they are not equivalent.
Considering the second interpretation, the first step in solving this integral would be using the algebraic identity : A^5 + B^5 = (A + B) (A^4 - A^3B + A^2B^2 - AB^3 + B^4), and then proceeding into a long chain of partial fractions.
The limit itself, is it e?
I suspect it's 3rd root of e..
Wow i never thought about this
Rigorous reasoning is inexistant in this video
Sir, you said that x = (1/3)t, but when substituting t in the limit,i.e., in x -----> infinity you should have been wrote it like (1/3)t -----> intinity. Why sir you wrote it as t -----> infinity
Dont worry about it
Wouldnt a = 1 and b = 1/3 for the last formula?? It comes to the same conclusion but i just want to know if im tripping
If you look at the equation after he subbed t for x, yes. But he did it for the original form of the equation. Remember t and x are just labels, they can be anything and don't actually matter -- they're only different so that we can tell them apart.
No. lim_t→∞(1 + 1/t)^t is not e BY DEFINITION, but by result. Don't they teach the binomial expansion any more?
(1+ a)^n = 1 + na + n(n-1)a^2/2! + n(n-1)(n-2)a^3/3! + ... Set n = 1/a = t and you get:
(1 + 1/t)^t = 1 + 1 + t(t-1)/t^2/2! + t(t-1)(t-2)/t^3/3! + .. As t→∞, we can see that t ≈ (t-1) ≈ (t-2), etc. Therefore t(t-1)/t^2 ≈ 1, and t(t-1)(t-2)/t^3 ≈ 1, etc. So we get:
lim_t→∞(1 + 1/t)^t = 1 + 1 + 1/2! + 1/3! +1/4! which is e _by definition._
There are two main definitions of e which are equivalent (and that's a theorem you can prove), one with the power series and one as the limit of the *sequence* a_n = (1+1/n)^n as n->infty. It's important to notice it's the limit of a sequence and not of the function (1+1/x)^x as exponentiation with real exponents is only defined through the exp function (basically after defining e and proving some properties)
Also, beware of "substituting" t(t-1)...(t-k+1)/t^k with 1 as, while it would be legal in a limit of a sequence by the theorem of substitution (not sure about the name, I learnt about it in another language) here you're considering an infinite sum, and as you probably know the infinite sum of infinitesimal errors can be non infinitesimal (well, in this case it still works, but you need to put more work to prove it)
@@notmymain2256 The most distinctive definition in my opinion is the base of logarithms whose anti-derivative is 1/x. But of course, there are lots of "definitions" that can be derived from each other. Nevertheless, it is rarely helpful to start from the limit as you suggest, since it is trivially found equivalent to the power series by use of the binomial theorem.
The limit is, of course, the same whether it is approached through the integers or the reals. Your point about exponentiation with reals is another reason why using limits is a less satisfactory starting point for a definition.
Why cant you just use logarithms?
Thank you Prof!
Compound rate.
Answer is on picture on right "e" or similar ;) (1+1/3x)^3x = e but how to aproxomate e^(⅓)…?
cool T-shirt, master!
Goddamn i asked this to my teacher and she couldnt reply now yt reccomends me this
More importantly, keeping aside how easy it is to see it is e^(1/3), can anyone tell my why it is tagged as differential calculus?
Because limits are taught in a differential calculus course.
Is 1/(infinity) indeterminate ?
No, it is 0. Indeterminate are like 0/0
Can anyone help me with simplifying e^(cosx)=-2sin(x)
have I passed calculus several years ago? yes. have I already knew the answer? yes. have I watched the video till the end because of the brilliant and easy explanation? also yes.
Thank you!
Once you learn the limit form of e this question is fairly easy to tell it wreaks of e😂
in fact, The Fact is a fact. It affects any student doing calculus and effective to solve such limits.
When I see such questions, the only thought I have: why did you decide that it can be replaced?
Practice and memory. Then you recognize patterns. "This looks like the formula for e, can I use that?". Some math youtube channels skip this part, and just say "let us write this a different way for no reason". That is not how it works. This channel is much better than most in explaining WHY you do those first interesting and sometimes creative steps. The rest is following the rules and seeing where you end up.
Rather than explain it in terms of indeterminate forms, I would say that lim f(x)^g(x) = (lim f(x))^(lim g(x)) only works if lim f(x) and lim g(x) both converge. In this case lim g(x) doesn't converge.
you are my hero
beautiful limit !
What happened to “my best friend”?
That is 1/(1-x) power series
I love this question.
306 / 5 000
hello, I love what you do however I would like you to add something more to your mathematics presentations such as the possible and applicable use of your equation in a specific field like chemistry, measurements, probability or related mathematics to our daily life, have a good day
Lhopitals rule;
Lim((3x+1)/(3x));
3/3 = 1
It's 1 power infinity type of limit so easy
why a=1/3? xd
1/3x = (1/3)/x
It's not 1^infinity = 1
when it is exactly 1^infinity yes it is one but here ,we dont have exact 1,we have 1+(like even 1.0000001^infinity = infinity) so cant just call it 1
Dis is on my shirt got me 😆
Give me this shirt. We needs this
Cool t-shirt 👕 !!!
Thanks!!
Nice short 😅