limit of x-root of x, calculus 1 tutorial

Рет қаралды 331,224

5 жыл бұрын

A classic calculus 1 limit problem, evaluating the limit of x-th root of x as x goes to infinity. This is an inf^0 indeterminate form example. We will have to convert the x-root of x to x^(1/x) and then write x as e^ln(x) and use L'Hopital's Rule. Try this next: what do you think what (infinity-infinity)^infinity could be? Answer here 👉 • (infinity-infinity)^in...
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Пікірлер: 536

@crosisbh1451 5 жыл бұрын

I'm really liking the whitechalkredchalk change right now.

@Jonathan-od5xc 4 жыл бұрын

It is most satisfying.

@LittleWhole 4 жыл бұрын

Jonathan Doster I agree. The sounds. Haha

@cubing7276 4 жыл бұрын

~blackpenredpen~ whitechalkredchalk yay!

@delusionaljackcoanfan7845 4 жыл бұрын

Fun fact: the absolute max of xth root of x occurs at x=e

@drahprub7750 4 жыл бұрын

I wonder how and why does it happen. Can you do the math?

@fiatfrenzy202 4 жыл бұрын

@@drahprub7750 www.wolframalpha.com/input/?i=plot+x%5E%281%2Fx%29+from+x%3D0+to+eulers+number

@orenjine3452 4 жыл бұрын

@@drahprub7750 (x^(1/x))' = (e^(ln(x)/x))' = e^(ln(x)/x)*(1-ln(x))/x^2 which equals 0 if x=e

@mauijttewaal 4 жыл бұрын

You misplaced the last bracket...

@drahprub7750 4 жыл бұрын

@@orenjine3452 intersting. Thanks!

@m3131m 5 жыл бұрын

I’m a retired engineer and high school math/physics teacher. Just want to tell you that I love your videos! Thank you so much!

@blackpenredpen 5 жыл бұрын

Thank you so much John. I love to receive comments like yours. It made my day.Thank you.

@arthurs5099 4 жыл бұрын

If you teach to do limit like this you should consider to resigne...

@tobiasgorgen7592 5 жыл бұрын

Pff i did this in my head and i am only 3 years old. But what can i say. I also have an IQ of -1/12

@blackpenredpen 5 жыл бұрын

: )))))

@pauljackson3491 5 жыл бұрын

I have an IQ of 100+1/12. I am just as smart as you are dumb. I kid. Normal IQ measurements, though they can't measure that small, is a normal curve so it is symmetric about 100. Of course you are talking about the joke of 1+2+3... = -1/12 but still. Using the error function to find the percentiles, an IQ of 0 is like 1 in 10e10.

@kimothefungenuis 5 жыл бұрын

@@blackpenredpen 1+2+3+4+...=-1/12

@inyobill 4 жыл бұрын

@@kimothefungenuis You replied to the wrong comment. I do that with distressing frequency. I have yet to see a valid justification for that conclusion, using KZfaq general Maths level Mathematics. Numberphile, for instance, uses sketchy Maths in an attempt to justify it, See Mathologer, and others. I have seen that it is just a Maths joke. Until proven otherwise, that is my assumption.

@darkseid856 4 жыл бұрын

@I Exist I don't think it's wrong . I mean it is wrong in the usual sense , but maybe it's something like complex numbers . Imaginary numbers don't exists in real life but the are still very useful for real life problems . Maybe the same way , getting a particular value from a divergent series is something special , something which is not practically correct but useful in an indirect way. But that's just what I think , I can be wrong .

@Nothing_serious 5 жыл бұрын

Clickbait. There's no red pen nor black pen.

@themeeman 5 жыл бұрын

blackboardwhitechalk

@brianjoelbasualdo7436 4 жыл бұрын

Jokes to you. I'm daltonic!

@necaton 3 жыл бұрын

@@themeeman red chalk: am i a joke to you?

@maxiomatic1609 4 жыл бұрын

physicists be like: lol infinity to the power of 0 is 1, problem solved

@imperialguardsman135 4 жыл бұрын

Can confirm, am a physicist

@jonatankelu 4 жыл бұрын

Yes, I wouldn’t think that’s an indeterminate form either as anything (except zero) to the power of zero is 1.

@dirktween244 4 жыл бұрын

@@jonatankelu Infinity Includes all numbers from 00 to beyond what you can fathom

@danielcordoba9018 4 жыл бұрын

@@jonatankelu Except in this case we have 0 as a limit, that's why it's an indeterminate

@dirktween244 4 жыл бұрын

Infinity to the power of anything, other than 0, is Equal to Infinity ! -- The larger the number, the more roots, before even getting close to 1 !

@larswassink2593 4 жыл бұрын

I really enjoy watching these videos. Most of them are just gimmicky videos explaining a tough mathematics problem, but they actually consist of standard maths solution steps which I think is really helpful. So far I've passed all my calculus courses at my university and it's all thanks to this guy. Really hope to see more of this (and maybe throw in another twelve hour differential equation marathon perhaps ;) )👌💪

@Tranbarsjuice 5 жыл бұрын

Blackpenredpen is looking fancy today :)

@blackpenredpen 5 жыл бұрын

Thanks!!

@carlosangulo2888 4 жыл бұрын

Blackpen, You are very handsome at this video. Congratulatons. All your videos have been excellent.

@Nxn908xxx 5 жыл бұрын

I can't impress a girl with this, cuz everyone knows it's 1.

@BlaqRaq 4 жыл бұрын

pusheen knowing and showing is a little different

@kummer45 4 жыл бұрын

I missed you Pen. Keep doing your great videos. Keeps me busy thinking about math problems. This should never go away.

@aliyardimoglu5629 4 жыл бұрын

I was algebra teacher in Richmond, CA.....that method of teaching, which is breaking the problem to the pieces and explaining them to the student body was really making a big difference on board.... congratulations to you, i hope one day you solve Putnam or such tests, in this explaining way, for people.

@jardelkaique2522 5 жыл бұрын

Fast and clear video. I liked it. Try to bring more in this format

@jsnmad 4 жыл бұрын

Love how you broke that problem down into the simple steps

@barakeel 4 жыл бұрын

Nitpicking: you didn't prove that the limit exists, you just proved that if it existed its value would be 1.

@bastiana.n.4277 5 жыл бұрын

Isn't he assuming the limit exists to take natural logarithm in both sides? What if the limit were to be negative, infinity or plainly non existent?

@arthurmeyre7214 4 жыл бұрын

Bastián A.n. Also why did he differentiate or take the ln ? for x > 0 x ^ (1/x) = e ^ (1/x * ln(x)) and ln(x)/x -> 0 when x -> +inf and e (x) -> 1 when x -> 0

@HariShankar-zy9ho 4 жыл бұрын

was going to comment exactly this

@DeepDeepEast 4 жыл бұрын

Yes he actually assumed it, but you can just define L=x^(1/x) and just continue...

@Lundburgerr 4 жыл бұрын

It's fine to make the assumption to see where you get from there. If a limit didn't exist we would see a contradiction at some point.

@lorenzorodriguez5703 4 жыл бұрын

@@arthurmeyre7214 think because he was already starting to use L'Hs rule but didn't state it. Also, can't have negative value in the natural log hence x > 0. Most importantly, he knew the answer so he didn't have to state why he used this method

@lutiga2226 4 жыл бұрын

Ok, the answer is 1... oh no, it’s 0.... hmmm c’mon, it may be 1.... no, 0 Then turns out it was 1

@marvingeheim8513 4 жыл бұрын

How could it be 0?

@adiarbiv4613 4 жыл бұрын

I just looked at it and knew it was 1

@PentaCorpStudio 4 жыл бұрын

Yeah, tbf 0 doesn't make much sense here

@lrkajighon2336 4 жыл бұрын

i thought the same way.

@benjaminbrady2385 5 жыл бұрын

Posted on my birthday! What a great gift :D

@blackpenredpen 5 жыл бұрын

Benjamin Brady thanks

@aayush3782 3 жыл бұрын

I just solved this exact same question today. Your way of teaching is really good even a amateur can understand well.

@abbamadoum4714 4 жыл бұрын

straight & comprehensive thanks sir

@RashadSaleh92 4 жыл бұрын

This is very professional and you know what you are talking about and you do not skip steps or oversimplify

@mohammadelsayed5715 4 жыл бұрын

Great work . Beautiful as usual

@hannahhagerty4532 8 ай бұрын

doing series tests right now and forgot how to take the limit of this kind of problem. thanks so much!

@raulfuerte5512 4 жыл бұрын

Well we took the scenic route, but still ended up in obviousville

@coryrea8914 4 жыл бұрын

Underrated comment lmao

@123SuperSomeone 4 жыл бұрын

That's some gorgeous math right there

@aepokkvulpex 10 ай бұрын

That dotted line at the end was impressive

@ibrahimabdullah9277 4 жыл бұрын

Looking forward for more , ty. Blesses ~

@marcow1011 4 жыл бұрын

short and simple explanation, i like

@felixjohnson4078 5 жыл бұрын

thank you! good video. I would love to see more tricky, tedious problems though. Maybe some famous integrals, or things like x^x^x!

@Hexanitrobenzene 5 жыл бұрын

Alex Mehregan He has done lots of these. Don't forget to check his older videos.

@JaiHall 4 жыл бұрын

He was killing that chalk 🤣🤣🙏🏿 Also, thank you. Very insightful.

@nanamacapagal8342 4 жыл бұрын

Did it in my head via a different method 4^(1/4) = 2^(2/4) 8^(1/8) = 2^(3/8) 1024^(1/1024) = 2^(10/1024) Note that as our x tends to infinity the exponent tends towards 0. So we have 2^0 = 1. Therefore x^(1/x) approaches 1 as x goes to infinity.

@rednassie1101 4 жыл бұрын

Me: Oh, anything^0=1 so that's probably true Wait, what is he.. What? ??? Why are there differentials all of a sudd... Ha, told ya it was (approached) 1

@user-jx4lf1bj2s 4 жыл бұрын

sometimes it is not thats why it's called indeterminate

@0upsla 4 жыл бұрын

With limits you always need to do the math, because the answer depends on the way the different terms approach infinity (or 0). There are even limits that are trully indeterminate, as they approach different limits depending on how you calculate them. (e.g. (-1)^x when x goes to infinity is indeterminate)

@IAmTehAg 4 жыл бұрын

@birolklp5574 4 жыл бұрын

Aloïs Christen I see what you did here, but come on that wasn’t an indeterminate here.

@dirktween244 4 жыл бұрын

Infinity to the power of x, is Infinity ! -- Only problem, is when x=0 or when x=Infinity ! ************* X^0 = 1 for all numbers, Except X=0 or X=Infinity Btw: What is (-1)^0 ? ************* ??? Does Infinity = 0 ?? >> Math says they are very similar, even to: -- Any equation with more than 1 of each, is invalid ! -- Actually need one to prove the other ! -- Both are "ultimate limits" ! -- Neither are True numbers ! -- They behave Identically ! >>Must include both, for language

@payammahbobi8029 4 жыл бұрын

Well done! Thank you!

@mr.9754 4 жыл бұрын

Actually, I don't think we need to use l'hospital's rule because x goes to infinity (significantly) faster than lnx, so we see the limit as x goes to infinity of lnx/x is zero (0)...but nice video DD

@pulakdas8998 5 жыл бұрын

Can you make videos about other inditerminate forms in limit problems?????????

@javierantoniosilva8477 4 жыл бұрын

Fun fact, ln(x)/x is the inverse (as a number) of x/ln(x); the aproximation of the counting prime function, done by Gauss (Prime Number Theorem). Since there are infinitely primes, one can informally hint that the limit of x/ln(x) tends to infty as x tends to infty, thus informally hinting that ln(x)/x tends to 0.

@javierantoniosilva8477 4 жыл бұрын

I cant recall if theres a relationship between lim f and lim 1/f. Probably just a coincidence, or Im overthinking, but its neat to see something number theoretic pop up here.

@CossZt6 4 жыл бұрын

Nobody: Me watching youtube at 2AM: oof a loopy limit

@carlosparedes7647 4 жыл бұрын

Spectacular!!

@ninnus12 5 жыл бұрын

Inf^0 = 1 i am done teacher

@antonyurkov15 5 жыл бұрын

WP GG not always, you can get e as well))

@gradecracker 5 жыл бұрын

Anton Yurkov What’s an rxample where you can get e from inf^0

@niconiconiiiiiiiiiiiiiiiii 5 жыл бұрын

@@gradecracker As inf^0 is an indeterminate form, you can get pretty much any value out of it. Example: (xa^x)^(1/x) -> a as x -> oo. (a > 1 or a = 1) So, plug in a = e, and you get: lim (xe^x)^(1/x) = e x > oo

@gradecracker 5 жыл бұрын

Rafael Nicolau so how can you get any value if we assign 1 to inf^0

@niconiconiiiiiiiiiiiiiiiii 5 жыл бұрын

@@gradecracker Dude, do you understand calculus? I mean, inf^0 is not equal to 1. It doesn't say at any moment it is. It's just that the limit as x goes to infinity of x^(1/x) give us the indeterminate form inf^0, and in this case, it approaches 1.

@nikolatesla6662 4 жыл бұрын

I like your channel very much

@justarandomweebyay5372 4 жыл бұрын

When chalk takes over black pen and red pen

@MaciejKlepaczewski 4 жыл бұрын

I think i spotted an issue with your proof. When defining L you've assumed that the limit in question exists and is a number. You can't just assume it, you need first to show that's true.

@MegaAgamon 4 жыл бұрын

You can just assume that L= x√x and pretty much it is the same proof

@MaciejKlepaczewski 4 жыл бұрын

@@MegaAgamon I don't see how that helps? Now you have lim L, then what?

@PentaCorpStudio 4 жыл бұрын

It isn't an issue because the limit is either in IR+* or +infinity, which you can technically transform in log(+infinity) = +infinity. The fact that he stays in ]0;+infinity] allows him to use the log.

@MaciejKlepaczewski 4 жыл бұрын

@@PentaCorpStudio I'm not suggesting that the result is wrong, I'm merely pointing out that this type of transition needs explanation why it's valid.

@respectpartii6302 3 жыл бұрын

@@MaciejKlepaczewski Probably if we assume we're calculating the limit at the extended real numbers field would be fine.

@Revanth292000 4 жыл бұрын

I like your style in this video. You look really great. :)

@Bhamilton-ws4go 5 жыл бұрын

Satisfying!

@Sosukz 4 жыл бұрын

New set up is cool

@simrich2375 4 жыл бұрын

I‘m a law student but I still enjoy videos like these since I always liked maths in school :D

@ethanbartiromo2888 4 жыл бұрын

It was a pretty intuitive answer, but the math makes it worth it!

@IsomerSoma 4 жыл бұрын

That's so cool that the n-root out of n limit to infinity converges to 1.

@roberttelarket4934 4 жыл бұрын

I like this!!!

@andrewhermit9098 4 жыл бұрын

Looking dapper ;)

@yoavshati 5 жыл бұрын

I'm never sure about my solutions for limit problems as I don't do them as often because we don't learn limits at school, and I managed to get the right answer

@SAHZ-xe3pz 4 жыл бұрын

Finally , solved on of your problems by myself

@markgraham2312 4 жыл бұрын

Awesome!

@georget8008 4 жыл бұрын

I have first noticed this, when i was in middle school and we had just been taught the square roots. I was choosing random positive numbers in the calculator and i kept on pressing the square root button. No matter what was my initial choice of number, finally I ended up with 1. At that point i did not know anything about limits and analysis. I doubt if we even had been taught algebra.

@HeyKevinYT 4 жыл бұрын

SAME

@nafrost2787 4 жыл бұрын

I hate to be that guy and nice story and all, but just for the record. If you choose some random number, and press the square root repeatedly, it's the limit of a^(1/x) where a is the random number, not x^(1/x), unless you press the square root button log(2,a) times.

@nostalgiafactor733 4 жыл бұрын

That's just how the calculator is rounding things off. You won't actually get 1 all the time

@yugalkishore8991 5 жыл бұрын

Really great KZfaq channel 👍

@tonk6812 5 жыл бұрын

Nice dots........like waltr lewin........😊😊😉

@hiccup3.14 4 жыл бұрын

The beginning looked like a mess But it's beautiful how the answer built up at the end as just 1 Man maths is amazing

@inyobill 4 жыл бұрын

Makes sense. For all x > 0: x^(1/x) > 1 AND x^(1.x) < (x + 1)^(1/(x + 1)). i.e., It is an infinitely decreasing series, with a lower bound of 1. Obviously I skipped some details.

@Hiltok 4 жыл бұрын

Or just consider x^(1/x) = x^(1)/x^((x-1)/x) which very obviously converges to x/x=1.

@dominicrowland9555 4 жыл бұрын

Awesome vid from benjamin

@mr.ketchup698 5 жыл бұрын

Thanks

@masheroz 5 жыл бұрын

If you do maths like a physicist, you say Lim(x->infty) ln(x)/x = 0 because x gets bigger faster than ln(x).

@n0ame1u1 4 жыл бұрын

ln(x) ∈ o(x)

@yvelkram 4 жыл бұрын

video : Solve question flawlessly somebody in earth : _tring to input 1, 2, 3..._

@hadar2win609 9 ай бұрын

the most impressive part is the chulkboard pen tickling at the end

@santanupatra993 4 жыл бұрын

Problems are so tough.But after you are done..It is pretty easy

@robertandersson1128 3 жыл бұрын

Wow, you're so fancy in this video! Chalks and suite, it makes me smile)

@pedroff_1 4 жыл бұрын

if you rewrite X as e^ln(X), you can get L = e^(ln(X)/X). X clearly grows faster than lnX, therefore lim ln(X)/X = 0 and e^(ln(X)/X) = 1

@Patapom3 5 жыл бұрын

Amazing!

@jackbunskin1892 5 жыл бұрын

How would you have to do this, if you aren't allowed to use l'hopitals rule?

@YuriyNasretdinov 4 жыл бұрын

The only issue that I have with the proof is that I think you can only apply functions to the inside of a limit not only when the function is continuous, but also when you first proved that the limit itself exists (and is finite). Other than that, thanks for a neat proof!

@float_geek 4 жыл бұрын

Hi, maybe I mistake but, I have the same conclusion saying that a large number (infinity is a number) to the power of zero is actually one, is not a indeterminate form right? Where I'm doing it wrong?

@bandamkaromi 5 жыл бұрын

I know but I'm watching to the end.

@nazirakhatun7637 Жыл бұрын

sir u dont have to solve full problem as u have mentioned there infiinity to the power is zero so here can we apply exponential rule that any number consist of power 0 is always 1 ?

@AceAufWand 4 жыл бұрын

The explanation is very clear but the only thing I miss in the explanation is the proof of L existence, when it is said that we call L the limit of this function to infinity, we imply that such a thing exists, which is not true for every function (cosinus). Is it a property of x to root x function ? In that case would the answer be more something like, if such a limit exists then it should equal 1 ? Till we proove that it exists.

@DemonSlayingFootball5208 4 жыл бұрын

I love using o(x) formulas for this

@jongyon7192p 4 жыл бұрын

x=2^u x^(1/x) = 2^(u2^(-u)) to inf, u*2^-u goes to 0, so answer is 2^0=1

@cmilkau 4 жыл бұрын

I know it's just a shorthand, but because this is so often a source for confusion: the symbol ∞√∞ is only justified if the two-input function (x,y) ↦x√y converges to the same value no matter which path you take to approach (∞,∞) in the x-y-plane

@rupendrayadav6056 5 жыл бұрын

Looking very cool

@blackpenredpen 5 жыл бұрын

rupendra yadav thanks

@johanlagardere7740 4 жыл бұрын

In first year of graduate we done that in our head you know easy peasy (but I didn't know about the multiplication dx/d ^^)

@adwait9806 4 жыл бұрын

Hey how can one see it ? I meant kind of geometry or intuition ?

@abcdefghijklm9697 4 жыл бұрын

In a bit more intuitive way you can say straightforward x^1/x = e^ln(x^x^-1)

@vicvic2413 4 жыл бұрын

Do the *SIMPLE* mathematical simulation (on Excel) and you will see that the function x ^ (1 / x) with increasing x first increases, reaches a maximum at x = e (number e), then decreases and asymptotically tends to 1. The correct answer: limit = "1"

@epicgamer-ur1wg 2 жыл бұрын

Cool

@vicvic2413 2 жыл бұрын

@@epicgamer-ur1wg:Let me add that the derivative of the function x^(1/x) has an extremum at the point x=e, and this proves that if b>a, then a^b > b^a, when a>e & b>e (3^4>4^3), and vice versa, if a

@tanaking5778 4 жыл бұрын

How can you say that you discovered de limit of ln(L) when you have only found out the limit of it's derivative?

@sahilraj3852 5 жыл бұрын

I am in a high school (preparing for the toughest engineering entrance exam) from India and I solved this question in just 4 steps !

@blackpenredpen 5 жыл бұрын

Good Job!

@sahilraj3852 5 жыл бұрын

Thanks @blackpenredpen for liking my comment and replying also. I just recently discovered your channel and solved some of the question from your videos and they are quite interesting. You are one of my fav maths teacher.

@farazalam3325 4 жыл бұрын

Finally a question I could solve in one go... (i am a doctor!)

@Zyx3ds18 5 жыл бұрын

I enjoyed this video very much! Keep up the good work!

@valentinbertrand3097 4 жыл бұрын

Une question bête pour un non initié, pourquoi la limite de la fonction est-elle égale à la limite de sa dérivée ??

@JonathonV 4 жыл бұрын

For once, both my predictions of the answer and of the method were right!

@tayserbinjafor1569 11 ай бұрын

We have dots of line at the end of the video.

@gerardogarduno8417 4 жыл бұрын

I was thinking and I guess that the idea of proving it with the ln is mere work. Someone pls correct me but I just wanna say, it doesn’t matter the number that the infinite gets, it could be 2 or 6262 or 286382637373 meanwhile the exponent is elevated to (1/x) and it approaches to infinite, the exponent will be 0, and any number elevated to 0 is equal to one.

@husklyman 5 жыл бұрын

Question: how to prove that lim(1+1/n)^n=e when n goes to infinity?

@embedded_ 5 жыл бұрын

This limit is equal e by definition of e number. If you want to prove that this limit exist and it is between two numbers ,you should use Newton binomial formula and geometric sum. Then when you'll find out that this limit of sequence is between 2 and 3 ,you could apply Weirerstrass theorem about convergence of a sequence

@MrBoubource 5 жыл бұрын

if you don't assume this is the definition of e, you can say that (1 + 1/n)^n = e^(n * ln(1 + 1/n)) and then use the first order approximation (taylor expansion ?) of ln(1 + x) near 0: ln(1 + x) = x + o(x) for x near 0. then, with x = 1/n, you get that ln(1 + 1/n) = 1/n + o(1/n) near infinity. hence, you get e^(n + ln(1 + 1/n)) ~ e^(n * 1/n) = e^1 = e. I hope it was clear. EDIT: Without using any expansion, you can deduce the limit from the derivative of ln(1+x) at x = 0, wich gives lim as x goes to 0 of ln(1 + x)/x = 1

@zackwoods5077 4 жыл бұрын

It's the definition of e.

@kekwait7959 4 жыл бұрын

Это практически очевидно, можно взять подпоследовательность 2^n It's almost obvious, u can just from start replace your x to 2^n (n approaches infinity), than ya'll get 2^(n*1/2^n). An' it's less than for example 2^(1/n), which definitely aproaches 1 when n goes to infinity

@redaiktane2837 4 жыл бұрын

But you can used,limit if x approach to plus infinity ln(x) over x approach to zero,but thank you tfor your effort

@KS-wy6ky 4 жыл бұрын

Do you really need the Lopetel's rule considering how simple the problem is?

@inordirection_ 4 жыл бұрын

Could someone be so kind as to explain why f(lim g(x)) is the same as lim f(g(x)) when 'f' is continuous function? (and why this isn't true when 'f' is not?) I'm referencing 1:20 in the video

@seagulyus9251 4 жыл бұрын

because the limit operator requires a continuous function to be well defined. if the function f you wish to operate on lim g with isn't continuous then the answer you get from f(lim(g)) could be different from lim(f(g)) due to f having different inputs in the two cases. however, if f is continuous then at any x the two examples will produce the same result. this is referred to as linearity iirc but specifically in operators. I'm sorry the actual term escapes me. but the long and short of it is that because they are continuous functions the order you apply them in shouldn't matter.

@BharathinfinityHegde 4 жыл бұрын

Let lim(x -> a) g(x) = c. The first case i.e f( lim(x -> a) g(x)) = f(c) However in the second case , lim(x -> a) f(g(x)) ; When x tends to a, g(x) tends to c. But f need not tend to f(c) unless it's continuous. To understand it better you can put g(x) = t So that, when x tends to a g(x) tends to c Or t tends to c The limit can be written as, lim(t -> c) f(t). By definition, when f is continuous at c, this limit equals f(c), otherwise it need to be equal to it.

@MurkoZawa 4 жыл бұрын

So it’s like saying that 1 to the infinity power approaches to infinity. But doesn’t it approach to e?

@inyouhead 4 жыл бұрын

I am from Russia, but accidentally found this video. I am very interested in such math. Thank.

@vicvic2413 4 жыл бұрын

Павел, видео хорошее, но товарищ -облажался- *ОШИБСЯ* ... Сделайте *ПРОСТЕЙШЕЕ* математическое моделирование (на Эксель) и увидите, что функция x^(1/x) при увеличении x сначала увеличивается, достигает максимума при x=e (число "е"), затем уменьшается и асимпотически стремится к 1. Правильный ответ: предел = "1" Правильное решение: kzfaq.info/get/bejne/d7R7bMypvbaad30.html

@cyberwindsl6735 4 жыл бұрын

Mind Blown... BOOM.....!!!!

@seeseefok7659 4 жыл бұрын

I love how you say ":)"

@flyfishjoren5283 4 жыл бұрын

It all seems so familiar and unfamiliar at the same time

@ludwigludwig3515 4 жыл бұрын

Unblievable. Sir, thank you, Sir.

@TheRoboticLlama 4 жыл бұрын

0:49 me: oh the answer is one

@Vnifit 4 жыл бұрын

No, it is an indeterminate form. Therefore, the limit could be anything. You cannot compute inf^0. Thus, the math is required to prove it. Just because it worked out to be one doesn't mean your hunch was right, it is dangerous to assume anything with indeterminate forms.