A limit that looks like e

[@___whateverr]

i solved it in my mind , and when i saw that you added ln i was so happy lol i felt like einstein thank you so much i love your channel

[@derrickbonsell]

You had me there with that n^2+n trick

[@samzii137]

Love from Pakistan 🇵🇰 Highly respected ❣️ To me you are a superman 🦸‍♂️ Thank you

[@meiwinspoi5080]

just love your teaching. its like a rockstar teaching. so unreal because you are yourself. un pretentious. i am a teacher and how i wish i could do as well as you do.

[@PrimeNewtons]

Thank you! 😃

[@Mr._Nikola_Tesla]

Sir, I'm going to go ahead and say that you are the most wonderful teacher I've ever come across. Also I have this problem from one of my old math challenge books (I already know how to do it, I just want to see your approach to it) It goes like- "We know that 1^0 = 1, so we can say, 1^0 = 1^1, since the bases are same, we can drop the base to get 0 = 1, find how to disprove this."

[@PrimeNewtons]

Nice question. Let me post it in the community.

[@Jon60987]

For the number 1, 1 to any power is 1. So the rule of dropping bases does not apply. The rule of dropping bases also does not apply to the number 0.

[@Mr._Nikola_Tesla]

@@Jon60987 You are correct, that is how I did it. But I believe there is a mathematical way to show this as well

[@Jon60987]

@@Mr._Nikola_Tesla Take the natural log of both sides. If the base is > 1, or between 0 and 1, then you can divide both sides by ln of the base to get the powers being equal. If the base = 1, then ln 1 = 0 and you can not divide both sides by ln 1.

[@Jon60987]

So here is my attempt to write it here: x^y = x^z implies that ln (x^y) = ln (x^z) implies that y*ln x = z*ln x, and if x is between 0 and 1 or greater than 1 then you can divide both sides by ln x to get y = z..

[@FedericoNassetti]

Keep going man i love your videos

[@PrimeNewtons]

Thanks, will do!

[@sandyjr5225]

Just curious, what's with the "Beware of dogs" caution at the end? It's a good caution though..

[@ThenSaidHeUntoThem]

Philippians 3:2 It is a verse of the bible.

[@Mr._Nikola_Tesla]

A little typo you have in the title.

[@PrimeNewtons]

Thank you. Fixed it

[@Orillians]

For a few days I was really bored and really did not want to watch your videos, but I watched one and now im binge watching all of the ones I missed lol

[@happyhippo4664]

I got an email that the T-shirts will ship next month. That is great.

[@Riazkhan12342]

Love from Pakistan 🇵🇰 ❤

[@Blaqjaqshellaq]

We now know what we'd only THOUGHT we knew...

[@PrimeNewtons]

That's an accurate way to put it.

[@nicolascamargo8339]

Muy buena la explicación, detallada y clara

[@vinayakmamtani]

It is the format of (1+f(x))^g(x) or 1^infinity so we write it as e^f(x).g(x) and then solve it as a normal limit

[@GreenMeansGOF]

My idea was to add and subtract n from the exponent but that may be more difficult.

[@xinpingdonohoe3978]

Mine too. Dealing with it may be more awkward since that n-√n doesn't divide too nicely into n²+n, to try and get more e stuff.

[@dean532]

Last time I took a peak into what you thought about Gamma functions. Back in the day, We expressed a couple of Vector Calculus results in terms of Γ(n) and incidently β (n,x) How about sharing some thoughts on the mighty Bessel functions too in maybe your upcoming videos..

[@epicmonke3319]

Ah hell nah that look at 3:28 got me putting my ears up

[@nothingbutmathproofs7150]

I am very surprised that that sqrt(n) didn't prevent the limit from being e. Nicely done.

[@jacksonsmith2955]

You shouldn't expect it to. With n -> infinity, the n^2 term will grow so much faster than the sqrt(n) term that it might as well not be there. Same with the n in n^2 + n. You only really need to pay attention to the part that grows the fastest when you have a limit going to infinity. Hell, you could take the limit of (1 + 1/[e^n + 2n^45])^(e^n + ln(n)^37) as n -> infinity and it would still be e, because the fastest growing term (e^n) trends towards infinity.

[@marcgriselhubert3915]

u(n) = [1 +(1/(n^2 +n)]^[n^2 + sqrt(n)] , so ln(u(n)) = [n^2 + sqrt(n)]. ln([1 + (1/(n^2 +n)]) We use equivalents when n goes to +infinity: [n^2 + sqrt(n)] is equivalent to n^2 and ln([1 +(1/(n^2 +n)]) is equivalent to 1/(n^2 +n) or 1/n^2. So ln (u(n) is equivalent to n^2/n^2 = 1 and limit (ln(u(n)) = 1 and finally limit (u(n)) = e when n goes to +infinity.

[@domanicmarcus2176]

At time 5:47, why did you look to your right towards the ground.

[@guidichris]

Nice!!

[@cesarcampuzanomartinez8182]

Can you solve the twin prime conjecture next please 😊

[@MrJasbur1]

Hey, completely unrelated question. But does anyone know if Prime Newtons will do a video on number theory or discrete logarithms?

[@PrimeNewtons]

Let me answer you here. The first 3 videos on this channel were on Number Theory. The audience for it is small. And you know, if they won't buy, don't sell is the the rule of KZbin. I will soon begin making exclusive long videos but only for channel members. So I can at least make videos that are less popular.

[@marcoghiotti7153]

Would it then be possible to generalise to any other similar form? I guess we could.

[@marcoghiotti7153]

I think as long as the 2 polynomials have the same degree you can generalise, as the lim of their ratio will go to 1

[@AlirezaNabavian-eu6fz]

Excellent

[@omerdvir1709]

I have a question. Could you take the limit to the base and the exponent in this case because you’d get the same answer but I don’t know if that’s just a lucky accident

[@boguslawszostak1784]

It is not true that the limit of the function is a function of the limit if the function is not continuous. The function x^a is continuous, so you can directly use this property without taking logarithms, which you just proved. If the continuous function g(x) has a limit y, then lim g(x)^h(x) = y^lim h(x).

[@lovishnahar1807]

sir this question was asked in indian statistical institute exam once

[@omograbi]

6:27 sqrt(n)/n^2=n^(1/2)-n^(4/2)=n^(-3/2)=1/(n.sqrt(n)) You've got this wrong here

[@nicolascamargo8339]

Así es, pequeño error

[@ruffifuffler8711]

Turned into the prostate wrench uvula function; (ln((l((1+1/(z^2+z))^(z^2+z^(.5)))^88), l being parametric.

[@michaelyu-jj8or]

this question can be solved by squeese rule

[@yiokreverso]

I have a challenge for you! to prove that lim x - > 2 of g(x) = { x^(2) if 0

[@johnpaterson6112]

Rather obvious that in the limit the n^2 will dominate the n and sqrt (n), so the answer is e.

[@adityashahi8184]

Well the another idea of this question was very impressive but there is another one .. e^lim (n²+√n)(1 + 1/n²+n -1) And then the answer was e

[@thabisonathanielmoremi8017]

Kindly check sqrt(n) /(n) ^2.