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!

[@johnalan8287]

My dear friend, after 100 years, people will still be watching your videos to learn calculus. You are so involved in your delivery, sharing and excellent in your concepts.

[@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!

[@FELISCATY]

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

[@kketaminekitty]

yay, finally managed to solve one of the problems in your videos on my own. thank you for your videos

[@carloruiz2228]

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

[@Charky32]

sir you explain it very simply, thank you

[@mitchelllevine5664]

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

[@xgx899]

For anyone who knows the subject, this is a one-liner: log(x^3+1/x^3)=log x^3(1+x^{-6})=3log x+1/x^6+o(x^{-12}). After dividing by x, this tends to 0. By continuity argument, the answer is 1.

[@ericabaez3033]

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

[@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?

[@Michaeladjei001]

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

[@farkliyahya]

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

[@PrimeNewtons]

Thank you so much 😀

[@FELISCATY]

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❤

[@BlueSiege01]

Thank you Sir for these videos!

[@GodwinSichone]

You have made my work easier sir 😁😉

[@mouraodomangua3488]

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

[@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

[@OnaMatsaseng]

This is so well explained! Thank you so much 😇

[@henry_dschu]

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

[@doctorb9264]

Clear and excellent job !

[@naklisahajgrover]

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

[@AhmedIsmail-z4i]

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

[@AnaGarcia-lh4mm]

the shirt is FYE!!! 🔥🔥

[@pianoplayer123able]

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

[@alpmuslu3954]

Love your work!

[@Khaidullah-q8y]

Your are the best sir

[@feedshark]

loved your infinity t shirt

[@nicolasb11]

WELL DONE PRIME NEWTON BRAVO !

[@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).😎

[@gkeic]

Where can I get this T-shirt?

[@PrimeNewtons]

Working on it

[@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.

[@vishalmishra3046]

Here 1 / x^3 tends to 0. so, result = [ x^(1/x) ]^3 = 1^3 = 1. (1 + zero) ^ infinity = e^constant but infinity ^ zero = 1 in the normal case of x and 1/x being infinity and zero. In this problem, powers of 1/x don't matter since they don't play any material role and all powers of x^(1/x) are powers of 1 = 1. So power of 3 also does not play any role. *General Rule* (X^n + X^-m)^(c/X) will tend to 1 for all positive integers m, n, c with X tending to infinity.

[@michaelbaum6796]

Very good example great👍

[@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)

[@MarSemedo]

great content, great guy!

[@PrimeNewtons]

Thank you

[@ЯНеДымок]

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 😂)

[@wolfix20021]

Never stop learning to infinity!!!

[@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

[@jannowak9052]

Koszulka jest świetna. Z chęcią zakupię.

[@kamumbai]

explained very well. Thank you.

[@dunerelaxtube4929]

Very elegant

[@sckani3432]

Nice, sir. S Chitrai Kani

[@novierjohnabdallahyousefebrahe]

You are doing well !!

[@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?

[@david-j8i4m]

Everyting raised to the power of 0 IS one 🎉🎉

[@eustacenjeru7225]

This is brilliant

[@surendrakverma555]

Very good. Thanks 🙏

[@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

[@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.

[@tomasbeltran04050]

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

[@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.

[@PauloDacosta-s1s]

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??

[@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

[@ISAACMAJEME]

Actually perfect.

[@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 ?

[@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)

[@BinethMinthusa]

🙏 Thank you Sir

[@lateefkareem]

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

[@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...

[@kaktusas1968]

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

[@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.

[@jharp49]

I liked your shirt.

[@hermannkengni]

but sir where is the natural log taken at the function

[@yogarasaponniah8586]

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

[@artandata]

I don't know but as soon as I realized that the expression is raised to 1/x, I thought in a very very large number raised to that power and to me the answer was 1 similarly as x tends to infinity the limit should be 1.

[@KhadidjaSebaa-t4e]

Wow it's so good

[@OpPhilo03]

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

[@capablancasqueen7574]

Sir, I'm in love with you!

[@PrimeNewtons]

Me too 😆

[@ВикторПоплевко-е2т]

11:50 almost like this limit ultimately has won

[@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

[@cliffordabrahamonyedikachi8175]

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

[@karimkemo5786]

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

[@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.

[@jamesmorton5017]

Anynumber to the power of 0 is 1?

[@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 ...

[@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?

[@rebtyxxz]

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

[@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 ?

[@北冥有魚-k2m]

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

[@chukwudisimere8463]

Beautifully done

[@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 ....

[@OludeleJacob]

I love this

[@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. 🤔

[@FrancisHealy-w9f]

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.

[@pelasgeuspelasgeus4634]

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

[@the_warpig1919]

All I can say is: Wow.

[@Tim-Kaa]

Nice

[@nikko2505]

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

[@JamesChigariro]

Nice

[@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.

[@rahleigh5829]

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

[@NwankwoBeloved]

Please am confused 😢

[@amdedemeke2544]

infinity to the power of zero is 1

[@maciejterakowski9062]

The result I found in 6 sec.

[@TheLukeLsd]

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