Limit at infinity of exponential function

[@junchen9954]

I feel like I'm watching Bobb Rosss but it's the math edition.

[@spudmcdougal369]

Hahaha Or Mr. Rogers does math.

[@JSSTyger]

"Over here we have some happy little limits."

[@JSSTyger]

Outstanding leszon.

[@jeffjones3287]

I love your calm, methodical, and understandable approach to teaching math skills! Thank you for the time and effort you put into these videos. these are great resources for students.

[@yoonsookim3979]

I agree :)

[@dougaugustine4075]

It's enjoyable to watch somebody take so much pleasure in doing and teaching math. It makes it so much more interesting!

[@cherryisripe3165]

I’m always pleasantly impressed by the love you have to teach mathematical skills, by your large scale knowledge field, by your calm and your pedagogy. By the way, I’ve always studied mathematics in French but I must recognise that I understand every subtility of your langage because your prononciation is agreeable and clear. Thanks so much for the effort you do to make mathematics so abordable and so easy.

[@PrimeNewtons]

Wow, thank you!

[@carloruiz2228]

Nice handwriting, clear explanation and a smooth pace. Nicely done!

[@QuantumCAT⁰]

I found your video randomly and found it that your teaching way is unique and started to watch regularly ❤

[@ayssinaattori9313]

Just found your channel, been loving the videos

[@Charky32]

sir you explain it very simply, thank you

[@mitchelllevine5664]

The world’s most lovable math teacher! The anti-Nash.

[@rcborges]

I was thinking of a more simplistic approach. As x goes to infinity the term 1/x^3 goes to zero and then the limit could be reduced to lnx^3/x, which translates to 3*lnx/x. As the lnx functions grows slower than any polynomial function the limit goes to zero and therefore the overall limit is 1. Is this way of thinking applicable?

[@BlueSiege01]

Thank you Sir for these videos!

[@ericabaez3033]

Gosh, this video is so refreshing. I love it.

[@QuantumCAT⁰]

We will never stop learning You never stop teaching ❤

[@arungosavi5698]

If this is a problem to be solved in an exam We will need you to print me at each step. Great solution with logical explanation. Love to learn more with you❤

[@cherryisripe3165]

Instead of using L’H rule, we could factor out x^3 inside (x^3+1 /x^3 and then use logarithm and the limite of lnx/x which is 0 when x approaches to the Infinity . So we get the same result.

[@pt3076]

Hi there, it has a much simpler solution, since as X approaches towards infinity the term 1/x^3, becomes 0, therefore it'll be simplified to find the lim of [(X^3)]^(1/x) when X approaches to infinity. That is equal to lim x^(3/x) , and in turn it is equal the lim of x^0 as X approaches the infinity, therefore it's equal to "1".

[@salvatorecosta875]

No! For istance lim(x→∞)((1+1/x)^x =e not equal to lim(x→∞)⁡(1)^x=1

[@jesemepardens9151]

​​@@salvatorecosta875The indeterminate form is 1^∞, but not ∞^0

[@wavingbuddy3535]

unfortunately that is an indeterminate form can't have ∞^0

[@farkliyahya]

ur shirt is definitly amazing i like it and your way of explaining

[@PrimeNewtons]

Thank you so much 😀

[@MichaelAdjei-up2ce]

What kind of teacher are you? 😮 Your teachings are always amazing ❤🎉 Keep it up.❤

[@shreyasjoshi5]

Is academic crush a thing? Cuz i think i have one of this guy

[@mouraodomangua3488]

Hello from Brazil, like ur channel very much I’m civil Engineering student 🤝

[@nicolasb11]

WELL DONE PRIME NEWTON BRAVO !

[@juliovasquezdiaz2432]

PROFESOR, UN GUSTO SALUDARLE. EN EL MINUTO 6.34 DERIVA EL LOGARITMO, PERO LA X DEL DENOMINDOR NO SE DERIVA ?. POR FAVOR SU COMENTARIO. AGRADECIDO POR LOS VIDEOS. SALUDOS DESDE PERU.

[@pianoplayer123able]

Can't believe that I solved that completely on my own and did everything right!

[@OnaMatsaseng]

This is so well explained! Thank you so much 😇

[@KahlieNiven]

from quick calculation by head, I also found limit = 1 ..in less than 5 seconds x^3 + 1/x^3 -> x^3 when x -> infinite (X^3)^(1/x) = x^(3/x) 3/x -> 0 and exponential part prevails upon exponented part so -> x^0 = 1 I agree, it's dirty. .. a good formal proof will always remain better.

[@doctorb9264]

Clear and excellent job !

[@user-pk4jl5wn4h]

Great explanation ❤ , please make videos on convergence and divergence of infinite series

[@РусскийПатриотЯша]

You mate are the reason I started to like calculus, and quite frankly you’re the reason I believe I can make it and become an engineer (although a manager one cause I still like money more than math 😂)

[@user-ru1vd2vw4m]

Your are the best sir

[@gkeic]

Where can I get this T-shirt?

[@PrimeNewtons]

Working on it

[@feedshark]

loved your infinity t shirt

[@diegosantosdeoliveira9929]

loved your approach man!btw at 6:25 when you applied L'hopital's rule and realized you needed to use some algebraic manipulation to evaluate the limit, if you kept applying L'Hopital 's rule until it was possible to evaluate the limit would you get the right answer?I mean, it's not that practical, but I'm just curious

[@alpmuslu3954]

Love your work!

[@henry_dschu]

Nice, like the way you do the maths, and your enthusiasm ❤🎉😊

[@jannowak9052]

Koszulka jest świetna. Z chęcią zakupię.

[@dunerelaxtube4929]

Very elegant

[@golddddus]

6:36 Instead of multiplying by x^4 it is simpler to immediately divide limit by x^2. Then lim x->inf (3-3/x^6)/(x+1/x^5)=3/inf=0. By the way, it avoids the confusion of the 0-0 sign. 4.26 The left limit is unnecessary. lim x->inf ln(y)= ln(y).😎

[@wolfix20021]

Never stop learning to infinity!!!

[@eustacenjeru7225]

This is brilliant

[@surendrakverma555]

Very good. Thanks 🙏

[@novierjohnabdallahyousefebrahe]

You are doing well !!

[@jharp49]

I liked your shirt.

[@kamumbai]

explained very well. Thank you.

[@tomasbeltran04050]

You may write e^y at ðe beginning so you don't forget. Just in case

[@AbouTaim-Lille]

The Ln function tends, if we can say, so "weakly" to the infinity, compared to X and actually the x is faster than ln |p(n,x )| for any polynomial of any fixed degree n. And for sufficiently large X , the inequality X> ln | p(n,X)| holds.

[@o-hogameplay185]

just a question: we can reorder it in a root format, where we get the y=inf'th root of (x^3+1/x^3). since y is not exponential, and (x^3+1/x^3)>=1 when x->inf, i can easily see, that the inf'th root of y is 1 when x->inf. or am i missing something?

[@WagesOfDestruction]

A simpler solution is to say as x-> infinite => y = (x^3) ^(1/x) = x^(3/x) from inspection as x-> infinite y=1, if you want to use L"H you can say y=x^(3/x) so ln(y)= (3/x) * ln (x) now do L"H and so you get (1/x) /1 or 1/x as x-> infinite ln(y)=0 => y=1

[@KazACWizard]

indeed a nice solution. however wouldn't the same logic be applied to what you said about the limit to the exponential? i rewrote the inside as e^(1/x)(ln(x^3+1/x^3)) and then put the limit into the exponent because exponential of base e if limit exists, exists for all exponents. then evalueate limit. i got zero too via the same method as you. then just replace that with e^0 and then of course you get 1.

[@michaelbaum6796]

Very good example great👍

[@lateefkareem]

Nice one. can you take the limit as x approaches 0? The result of that is more fun and unexpected

[@user-id5do9ly3z]

Can you not solve this limit without use L’Hospital? It is strange because I thought that limit was developed before limit so how Euler solve determine his number??

[@yogarasaponniah8586]

Substitute a large valve for x ,Example x=1000 You will get the answer 1

[@BinethMinthusa]

🙏 Thank you Sir

[@gamingstudio7103]

I transformed it into and exponential and then I neglected the 1 before the x^6 and I got the same result. Is it right ?

[@peta1001]

Quite often mathematics gets to a solution by applying a convenient (not necessarily true) assumption. Here it seems to be an assumption that (x) = (x+3) when x becomes infinite. If you imagine that any weight that we cannot express, measure or weigh is the same or equal to infinite, we make a mistake. A heavier rock is always heavier than a lighter one, no matter how small the difference in weight is.

[@user-cb3yw2ju7w]

Everyting raised to the power of 0 IS one 🎉🎉

[@user-zy3eo4uf2e]

Wow it's so good

[@capablancasqueen7574]

Sir, I'm in love with you!

[@PrimeNewtons]

Me too 😆

[@kaktusas1968]

Don't we immediately see that a parenthesis to the zeroth power will be equal to one?

[@meditatingcow5113]

Hey can you have a look at this lim(x->♾️)[4^n+5^n]^1/n ?

[@PrimeNewtons]

I like this

[@meditatingcow5113]

@@PrimeNewtons then how about a video on it

[@cliffordabrahamonyedikachi8175]

Simply substitute x as infinity.knowing that infinity to the power of 3 . It equals( 2)^1/1. This is 2.

[@OludeleJacob]

I love this

[@pizza8725]

I knew that it would be 1 bc x³+1÷x^3 when it's infinity is really just x^3 and bc 1÷x would outgtow x^3 really fast so it's really just x^1÷x and that is 1(when x is aproching infinity)

[@hermannkengni]

but sir where is the natural log taken at the function

[@karimtarek5786]

But u can tell that limit approaches 1 from the exponent. Because the exponent is 0.

[@OpPhilo03]

I am solve limit any questions by Infinite Countability 😊. Some times I am wrong but most of the time my right.

[@WePhFr]

If I draw a graph of that function I see the value of 2 as the limit goes to infinity. Where is the mistake?

[@WePhFr]

I was wrong - must go farther than some hundred to see the Limit of 1...

[@user-ib8bh4jq2v]

Nice

[@josleurs4345]

why do we not use just a calculator or spreadsheet , just to get a feeling ..., that is the way I explain it to my children ...because de l hopital is tricky because of the conditions of it ...

[@jamesmorton5017]

Anynumber to the power of 0 is 1?

[@josleurs4345]

I would say that it can be easier ... ( X^3 + 1/X^3 ) ^(1/x) the second term in the sum goes to zero ... so left ( x ^ 3 ) ^( 1/x ) ... x ^ ( 3/x ) and then ln en de l hopital is easier ....

[@user-gq2oj2nk4h]

Using ln(ab) and L.H. , gives lnY=0, and so Y=1 as x→∞

[@Anmol_Sinha]

How did you use Lhopital? It only works when you have a 0/0 indeterminate form right?

[@Anmol_Sinha]

But we will still get 1 if we use the property of logs + it's taylor expansion to solve it

[@PrimeNewtons]

I used to think so too, but 0/0 is also inf/inf. So it works for both. Never Stop Learning!

[@Anmol_Sinha]

@@PrimeNewtons cool! And thanks

[@junchen9954]

It's not like it only works for 0/0 or inf/inf but these are the only two situations for which the differentiation would be necessary. If either the denominator or the numerator was neither zero or inifinity then you would have already got the result you were looking for, so you differentiate the expressions on both sides till you get a result as such. E.g if the denominator was zero and the nominator wasn't then clearly the limit would be infinity so there would be no need to further differentiate the equation. You wouldn't need to differentiate an equatoin whose target limit is 2/0 coz you would know the limit would be ∞. L'H rule works for all cases except the target result only practically means something to you when the target value isn't 0/0 or inf/inf.

[@Anmol_Sinha]

@@junchen9954 uhmm.. the derivation of lhopital comes by assuming a 0/0 form (or inf/inf). So I don't think that it will apply to other cases. (The proof is in 3b1b calculus playlist if you are interested)

[@yvngrxxd9078]

i just learned limits 2 days ago so someone please tell me if i did something horrendously wrong 💀💀💀💀: let y = x^3 + 1/x^3 so lim x -> ∞ (y)^1/x so x for infinity is a number that keeps getting bigger and bigger and (y)^0= 1 so the answer is just 1?

[@imshiruba]

The y approaches infinity which leaves us with an indeterminate form of limit, infinity^0. But honestly tho, you did pretty good for the first 2 day of limits

[@user-lu9fg7pc9q]

11:50 almost like this limit ultimately has won

[@the_warpig1919]

All I can say is: Wow.

[@user-ud1zv2yh3r]

This is an example of a incredibly complicated description of a trivial result. The "+1/x^3" could not possibly make a difference since this quantity goes to zero. The problem reduces to proving (x^3)^(1/x) goes to zero as x goes to infinity. Taking logs we get (3/x)log(x). Everyone knows that Log(x)/x goes to zero as x goes to infinity. So this could be proved in about 30 seconds.

[@angelavitaliano5200]

Occorre specificare che la variabile reale x ,tende a + infinito ; la funzione a valori reali di cui si vuole calcolare il limite , non è definita per tutti i valori di x negativi.

[@chukwudisimere8463]

Beautifully done

[@gabonviper5426]

Isn’t it obvious from the beginning, that the limit is 1? If you look at the initial expression, it is clear that it is like “something” to the 0 degree. So, what ever is “inside” the brackets, whole expression is in zero degree. Anything in 0 degree equals 1. 🤔

[@265user]

Is it wrong to just raise infinity to zero and say its 1? Since any numbers raised to zero is 1

[@Kraken-lm1cx]

Nah, infinity is not a number. Also the case of anything to the zero being 1 isn't always true as 0^0 is a counterexample

[@helm36]

What?0^0 is 1, it's not a counterexample

[@Kraken-lm1cx]

0^0 is not one always. If you look at the function x^0 it seems 0^0 will be one but if you look at the function 0^x it seems to the value will be 0@@helm36

[@helm36]

@@Kraken-lm1cx function x^x as x approaches 0 equals 1. For example you take 10e-10 to the power of 10e-10 it will be pretty close to 1

[@helm36]

@@Kraken-lm1cx yeah 0^x as x approaches 0 is not 1, but 0^0 being 1 is basically a math rule due to the reason I mentioned above

[@thaerthaer1120]

There is no need for all of this, generally anything, to power zero equal to one ,so from the beginning, we have infinity to power zero, which is equal to one

[@PrimeNewtons]

Are you sure?

[@thaerthaer1120]

@@PrimeNewtons why not ?

[@pelasgeuspelasgeus4634]

Try graphing it desmos and you will see that the limit is 0 and not 1.

[@maciejterakowski9062]

The result I found in 6 sec.

[@nikko2505]

It is much easier to do without L'Hopital's rule. Through the sum of cubes

[@amdedemeke2544]

infinity to the power of zero is 1

[@rahleigh5829]

Let x = 99999999 And you will get 1.000000553 Approx. = 1

[@toto-yf8tc]

Dude, ln(x^3+1/x^3) ~3ln(x) at infinity. Period. Stop with the senseless computations

[@NwankwoBeloved]

Please am confused 😢

[@lucid6392]

I mean the lim as x-> infinity of the exponent (1/x) is 0 so you could immediately see it was 1

[@chukwudisimere8463]

Not really... At first glance, if we look at it the way you suggest, it gives an infinity to the power of zero which is an indeterminate form

[@TheLukeLsd]

let k= 1/x so the L= lim k--> 0 (k³+1/k³)^k = 1.

[@eiseks3410]

The dislike button is not enough

[@rushhourgaming]

Ok let me teach you shortcut to find limit to solve easily any question For infinity put x=999999 digits as much you want For 0 put 0.00001 more digits more accurate answer same for other numbers Now if we see this question (9999999³+(1/999999³))^(1/999999) So it becomes (9999999999......)⁰ As we know a⁰=1 Thus answer is 1 This trick almost work for all questions and if it's MCQ question it's fastest way to find your answer

[@Ichigo-gp9vq]

I'll make sure to use this in exams

[@austinpowersasmaozedong]

of course this is great for mcqs but when it comes to a 6 marker written method then you will unfortunately have to learn how limits actually work 😂

[@265user]

He said infinity raised to zero it's undefined... Can someone clarify this am lost

[@rushhourgaming]

​@@265userwell infinity also no so anything power to 0 is 1 But in case of 0⁰ there are different answer Like anything power to 0 is 1 0 to power anything is 0 so it's undetermined

[@rushhourgaming]

​@@austinpowersasmaozedongyaah but you can use it to cross check your answer But in MCQ it is very helpful

[@coolgameplays1454]

easy

[@coolgameplays1454]

just get rid of 3/x^4 and 1/x^3 beacause x goes to infinity then theses fractions is 0 then you can simplify by 3x^2/x^3=3/inf=0

[@ralfdetemple6996]

This ist totally wrong. The Expression converges to 1 and not to 0.

[@PrimeNewtons]

Where did you find 0?

[@sergzerkal1248]

В конце лишнее и не рационально.

[@jerapahkul]

Dimas hotwil

[@nickmahardika2831]

Jono matchbox

[@jerapahkul]

👲

[@user-nd7th3hy4l]

lim= 0 X->infini

[@user-ud1zv2yh3r]

You make trivial math seem complicated.

[@PrimeNewtons]

Not my intention. I wish I had your insight.

[@ghamoz]

Troppi passaggi il risultato è già evidente alla prima applicazione di hopital ( e comunque si vede qnche prima)