If the limit exists, you can take the limit into the exponent because f(x)=e^x is a continuous function. First show the limit exists and then use the continuity of e^x.
Precise, clear, well presented, well decomposed +elegance and + well spoken english (I'm not a native english teacher). You are definetly one of my favourite math youtuber ! :)
...Good evening Newton, I hope your doing well. You look like a "mathematical" alchemist, who manages to tackle every seemingly unsolvable limit! Now, only the easiest thing left for you, namely to create gold out of anything! (lol). Another great performance with visibly a lot of enthusiasm. Thank you Newton, Jan-W
Well cant you rewright this as 1÷x^1÷x and by ising l hopital rule,you will the limit as x apriches infinity of 0÷((-x^1÷x-2)(ln(x)-1))=0÷(-x^1÷x)ln(x)×x² and bc x^1÷x aproches 1 and bc x² grows much faster than ln(x) than that expresion aproches 0 so you will get 0÷0 and in limit standards it equals 1
6:20 there's another proof that we use in class that is the limit of a function/x tell us its parabolic branch direction , if it follows the vertical axis it'll be infinity, and if it follows the horizontal axis it'll be 0
Lol I did it by intuition and I was correct. My way of thinking was that we have something raised to the power of something that is close to 0 and therefore the answer is 1. But number e can also come in handy here. Imho you overcomplicated a bit with introducing logarithm and y, it can be solved faster If you know the formula for e: (1+1/x)^x when x tends infinity is e. Just transform initial equation to this and you will essentially get the same result.
Not all limits are taught before derivatives. Some limits require more sophisticated strategies. That's where L'Hopital's rule comes in. There also other strategies.
In algebra, 0^0 is defined and = 1. analysis workers don't have the same opinion. as we use limits here, we not define it. However, (1/x)^(1/x) -> 1 still. (0^0 been controversial for 2 centuries and still is) to me it's as much defined as 3x1/3 = 1 and not 0,99999999999999999... Edit : tho what happens when x < 0 ? -normally not defined in analysis because of ln- ? (ie. (1/-0.5)^(1/-0.5) => -2^(-2) => 1/(-2)² = 1/4 thus, is defined.
8 minutes to solve an elementary exercise that can be solved in few seconds! Changing variable y = 1/x the limit is reduced to the limit as y goes to 0^+ of y^y, that is exp(ylog y). Now ylog y goes to 0 and so y^y goes to 1.
If you have to do it by observation anyway after doing all that math, then why can’t you do the same in first step, 1/x to power of 1/x is zero to power of zero which is equal to 1?
My logic was a bit more dumb. I just said , ok , its like 1/(x^(1/x)) this is continuous , it can't do to many jums, so lets just find n^(1/n). But this is much more elegant
Только у вас постояннпя ошибка , когда вы пишитп логарифм, с именно основание не указано. Log имеет любое основание хоть 2, хоть 10, хоть е . Поэтому десятичный логарифм имеет запись Lg= log(10)
The limit of that function is zero. and if you check the numbers you can use, you need to know: X is an element of the real numbers, but without zero. if you look left and right from zero, it goes to plus infinity.
The base approaches ∞ and therefore you have the indeterminate form ∞^0. That's why you just can't calculate the limit when the function is written as x^(-1/x).
@diegocabrales since the exponent gets smaller and smaller, the fact that it's negative is less and less significant, meaning the base can be considered approaching infinity
@@jamesharmon4994You're right that the base doesn't approach 0 but ∞-my mistake. I've already edited my comment. But that just only changes the indeterminate form: now we have ∞^0 one. The thing is that the result is different depending on which are the paths that both functions take to approach 0 and ∞, so you just can't say that ∞^0 = 1 because it could be convergent to another number, or diverge to ∞, or -∞, or even not be defined.
@@jamesharmon4994 Yes, you're right, n^0 = 1 for every number n ≠ 0. If n = 0 then we have 0^0, which is undefined. But ∞ is not a number, so you can't apply it to ∞. ∞ is a limit approach and ∞^0 is an indeterminate from, meaning it could be any number, or diverge to ∞, or -∞, or even not be defined. All depends on how do our functions approach ∞ and 0, respectively. That's why you just can't say that ∞^0 = 1 and be ok with that. Let me use another indeterminate form to understand it. You could say that since all real powers of 1 are equal to 1, then 1^∞ = 1. However, let me introduce you to e, Euler's number: e = lim(x --> 0)(1 + x)^(1/x) Clearly the base approaches 1, and 1/x approaches ∞. However, 1^∞ = e ≠ 1. But we can have 1^∞ = 1, like in this example: lim(x --> ∞)(1 + 1/2^x)^x = 1 And we can continue doing that with two functions which make the indeterminate form 1^∞ and be equal to any number, or diverge to ∞, or -∞, or even not be defined. Then what you apply to numbers can't be applied in general to limits.
In my very humble opinion, the proof here is wrong. The limit at infinity of 1/X to the power of 1/x is 0 to the power of 0 (as the limit at infinity of 1/X is 0). But 0^0 has no definite agreed value (only non zero number to the power of 0 are equal to 1). I suspect that replacing the limit of the function by the function of the limit is a trick that does not work if the limit does not exist. So still in my humble opinion, the proof works because you furst define and accept that 0^0 is defined.
@@geraldvaughn8403My mistake. I meant that he accepts that x^0=1. He said that 0^0 is not 1 but e^0=1. First to be clear, I know what official academia says. It's a convention that x^0=1 but it never stood with me because the exponent definition is clear and it says that a^b means "take 0 and add a multiplied with itself b times". Now don't get confused. Actually multiplication happens b-1 times but you get the point. So, if b=0 what would be the logical result? 0 or 1?
If the limit exists, you can take the limit into the exponent because f(x)=e^x is a continuous function. First show the limit exists and then use the continuity of e^x.
Precise, clear, well presented, well decomposed +elegance and + well spoken english (I'm not a native english teacher).
You are definetly one of my favourite math youtuber ! :)
Awesome videos and awesome teacher! Thanks a ton for doing these. 😃
So comprehensive!! Thanks sir. Greetings from the Philippines!
Just love your way of teaching.😊😍🇧🇹- Bhutanese student!
...Good evening Newton, I hope your doing well. You look like a "mathematical" alchemist, who manages to tackle every seemingly unsolvable limit! Now, only the easiest thing left for you, namely to create gold out of anything! (lol). Another great performance with visibly a lot of enthusiasm. Thank you Newton, Jan-W
Prime Newtons always delivers! 😃
A fine teacher indeed!
Although the 0^0 is not generally defined, it is defined as 1 in a special case, so it can be intuitively considered as 1.
Very clear again, thanks !!
Love from India ❤🇮🇳
고등학교 때 배우던 내용을 40년 지나서 다시 보니 재미있네요
Well cant you rewright this as 1÷x^1÷x and by ising l hopital rule,you will the limit as x apriches infinity of 0÷((-x^1÷x-2)(ln(x)-1))=0÷(-x^1÷x)ln(x)×x² and bc x^1÷x aproches 1 and bc x² grows much faster than ln(x) than that expresion aproches 0 so you will get 0÷0 and in limit standards it equals 1
I love watching these because they are such a relaxing way of thinking through math.
Неплохо. Отличный стиль ведения уроков, бро!
6:20 there's another proof that we use in class that is the limit of a function/x tell us its parabolic branch direction , if it follows the vertical axis it'll be infinity, and if it follows the horizontal axis it'll be 0
You can let y=1/x, then the limit is equal to: limit(y^y)
When y-->0. This is a well known limit in calculus, and it's equal to 1.
This is a much cleaner solution
@@herryeric454 nods but why it is a well known solution. that's what the video answers to.
Very good. Thanks 🙏
Hope you are doing well my favourite teacher thanks so much🥰🥰
The log of the function is -log x/x ->0 for x->infty. Hence the limit of the function exists and equals to 1.
"That's all folks!"
Molto bravo Professor!!!
Amazing your vídeos. Just a sugestion , plot graphical function output also. Tanks😊
How about just make the limit in the form of y=lim n->0 (n^n) and take the log of it with base of n, then you’ll find n^0=y
Wouldn’t this be the same as
The limit of (n root(1/n)) as n approaches infinity?
Best teacher !!!!
+1 subscriber
Amazing thank you so much
We can make a change of variables. Let y=1/x. The limit of y^y as y approaches 0. The limit is then 0z
Did this in my head. Knew there was gonna be a log/e with limit
great job prof.
Thank you
By the way, if you use lg instead of ln, you get 10⁰, which of course is also 1.
thanks man!!!👋
I'm fact it was intuitive as it seems to be something like 0^0 which is 1
Lol I did it by intuition and I was correct. My way of thinking was that we have something raised to the power of something that is close to 0 and therefore the answer is 1. But number e can also come in handy here. Imho you overcomplicated a bit with introducing logarithm and y, it can be solved faster If you know the formula for e: (1+1/x)^x when x tends infinity is e. Just transform initial equation to this and you will essentially get the same result.
пардон, а какой тип непррывности? поточечная? равномерная?
The date when this comment was posted, Prime Newton's had 1/5√2 million subs. Good going 👍
Haha 😄
Professor, there is one thing I don't understand, límits are taught before derivatives so why do I use L'Hopital in límits?
Not all limits are taught before derivatives. Some limits require more sophisticated strategies. That's where L'Hopital's rule comes in. There also other strategies.
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The intro song is pretty cool and relaxing. 😂
Thanks to @Kayomusiq
bangin vid
In algebra, 0^0 is defined and = 1.
analysis workers don't have the same opinion.
as we use limits here, we not define it. However, (1/x)^(1/x) -> 1 still.
(0^0 been controversial for 2 centuries and still is)
to me it's as much defined as 3x1/3 = 1 and not 0,99999999999999999...
Edit : tho what happens when x < 0 ? -normally not defined in analysis because of ln- ? (ie. (1/-0.5)^(1/-0.5) => -2^(-2) => 1/(-2)² = 1/4 thus, is defined.
let 1/x = a, as x approaches infinity, a goes to zero. Now the limit becomes a^a as a goes to 0 which ends up in 1.
Perfect expression..
Thanks
subscribed 📈
8 minutes to solve an elementary exercise that can be solved in few seconds! Changing variable y = 1/x the limit is reduced to the limit as y goes to 0^+ of y^y, that is exp(ylog y). Now ylog y goes to 0 and so y^y goes to 1.
This is not meant for math olympiad participants. The problems are the fodder for giving us a feel for math.
If you have to do it by observation anyway after doing all that math, then why can’t you do the same in first step, 1/x to power of 1/x is zero to power of zero which is equal to 1?
why dont you just use the squeeze theorem
this is like saying the root of 1/x, or 1/limx->infinity of x
My logic was a bit more dumb. I just said , ok , its like 1/(x^(1/x)) this is continuous , it can't do to many jums, so lets just find n^(1/n).
But this is much more elegant
Thanks for an other video master
My pleasure!
👏👏
when limit x tends to infinity in equation (1/2)^ x then the answer is zero. How it is possible?
1^x is always 1 , while 2 gets bigger , and bigger as we head to infinity so it's gonna be 1/(2^x) so the lim is equal to 0
My guess is that it's one since 1/x ^ 0 = 0, if you just ignore that 1/x = 0 itself. Not rigorous though.
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Только у вас постояннпя ошибка , когда вы пишитп логарифм, с именно основание не указано. Log имеет любое основание хоть 2, хоть 10, хоть е . Поэтому десятичный логарифм имеет запись Lg= log(10)
The limit of that function is zero. and if you check the numbers you can use, you need to know: X is an element of the real numbers, but without zero. if you look left and right from zero, it goes to plus infinity.
no. if you want a better way to calcul this limit you can just define X=1/x and so you'll have lim_X->0 (X^X). as you probably know 0^0=1
@@nhazeg2344 0^0 is undefined
@@bakashironot always, in this case it's intuitively 1
@@scott5388 i know that
Thanks Good day 8:13
8:13 l like it
The lowest value it gets is exactly e
Why couldn't you take the limit as x approaches infinity of x^(-1/x)? Here, it's clear that -1/x approaches 0, therefore the limit is x^0... aka 1.
The base approaches ∞ and therefore you have the indeterminate form ∞^0. That's why you just can't calculate the limit when the function is written as x^(-1/x).
@diegocabrales since the exponent gets smaller and smaller, the fact that it's negative is less and less significant, meaning the base can be considered approaching infinity
@@jamesharmon4994You're right that the base doesn't approach 0 but ∞-my mistake. I've already edited my comment. But that just only changes the indeterminate form: now we have ∞^0 one. The thing is that the result is different depending on which are the paths that both functions take to approach 0 and ∞, so you just can't say that ∞^0 = 1 because it could be convergent to another number, or diverge to ∞, or -∞, or even not be defined.
@@diegocabrales to my knowledge, there is only one number when taken to the zeroth power that is NOT 1 is... zero.
@@jamesharmon4994 Yes, you're right,
n^0 = 1
for every number n ≠ 0. If n = 0 then we have 0^0, which is undefined.
But ∞ is not a number, so you can't apply it to ∞.
∞ is a limit approach and ∞^0 is an indeterminate from, meaning it could be any number, or diverge to ∞, or -∞, or even not be defined. All depends on how do our functions approach ∞ and 0, respectively. That's why you just can't say that ∞^0 = 1 and be ok with that.
Let me use another indeterminate form to understand it.
You could say that since all real powers of 1 are equal to 1, then 1^∞ = 1. However, let me introduce you to e, Euler's number:
e = lim(x --> 0)(1 + x)^(1/x)
Clearly the base approaches 1, and 1/x approaches ∞. However, 1^∞ = e ≠ 1.
But we can have 1^∞ = 1, like in this example:
lim(x --> ∞)(1 + 1/2^x)^x = 1
And we can continue doing that with two functions which make the indeterminate form 1^∞ and be equal to any number, or diverge to ∞, or -∞, or even not be defined.
Then what you apply to numbers can't be applied in general to limits.
Why don't you Just say
e^(ln(x)/-x)
= 1/√(exp(ln(x))
= 1/√x ?
Thats supposed to be the xth root sorey
bro just use lhopitals rule before applying the limit
In my very humble opinion, the proof here is wrong. The limit at infinity of 1/X to the power of 1/x is 0 to the power of 0 (as the limit at infinity of 1/X is 0). But 0^0 has no definite agreed value (only non zero number to the power of 0 are equal to 1).
I suspect that replacing the limit of the function by the function of the limit is a trick that does not work if the limit does not exist.
So still in my humble opinion, the proof works because you furst define and accept that 0^0 is defined.
This limes is basically 0⁰. But while 0⁰ itself is undefined, this limes is 1, as x⁰ = 1.
lim t -> 0+ t^t is not undefined, it's equal to 1
Lim=1
So, your basis is that 0^0=1. Right? Your methodology is valid but it doesn't feel right...
He never said that. In fact he said that was undefined
@@geraldvaughn8403My mistake. I meant that he accepts that x^0=1. He said that 0^0 is not 1 but e^0=1.
First to be clear, I know what official academia says. It's a convention that x^0=1 but it never stood with me because the exponent definition is clear and it says that a^b means "take 0 and add a multiplied with itself b times". Now don't get confused. Actually multiplication happens b-1 times but you get the point. So, if b=0 what would be the logical result? 0 or 1?
One
@@geraldvaughn8403 OK. Could you elaborate?
@@geraldvaughn8403 What's the matter? Can't you explain why you find reasonable nothing be equal to 1?
Do the math sugestive ok
0^0 approaches 1
@ pbbperera