What is the biggest n-th Root of n?

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Today we investigate what the biggest x root of x (x^1/x) is and how it behaves in the limit. Enjoy :)
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30 янв 2022

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Комментарии : 277

@PapaFlammy69 2 года назад

*_Ayo fapers, thx for watching

@tridivsharma2342 2 года назад

Flammy be looking like a bodybuilder lol

@cicik57 2 года назад

just take the derivative (1-lnn) *n^(1/n-2) and compare to zero. The extremum is at lnn=1 : if lnn < 1 then dervative is > 0 - accending, if lnn>1 then descenting. so the extremum is at n = e = 2.71. Since this number is clother to 3, for natural n it shoul be 3 check : ∛3 = 1.44, √2 = 1.41 , ∜4 = 1.41

@AdityaYadav-vn6kc 2 года назад

Solve a JEE ADVANCE question if u are not alien

@oni8337 2 года назад

@@AdityaYadav-vn6kc Jens already said he is tired of people shoving India in his face

@cicik57 2 года назад

@@AdityaYadav-vn6kc whats that?

@AndrewDotsonvideos 2 года назад

Fun fact about Jens, he actually never learned to read. Good job man you make us all proud!

@Gameboygenius 2 года назад

Amazing. Like if u cry evertim.

@glory6998 2 года назад

Hello Dotson bro

@theproofessayist8441 2 года назад

Yes can't believe this mad lad hates to "read" and got this good at mathematics without reading any textbooks. All rigorous skimming of lecture notes and PDFs and still types out professional looking LaTex PhD papers. Funky bullshit powers activate!!!

@soham9119 2 года назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-BmXwvWnokz4.html

@melodyparker3485 2 года назад

Wuck

@decare696 2 года назад

for the log limit just notice that log(n)/n = -log(1/n)/n. Then substitute u=1/n and you get lim(u->∞) -u log(u), which is easily seen to be -∞.

@PapaFlammy69 2 года назад

nice!

@ezranwobodo4912 2 года назад

Why would you take the limit as u -> \infty when the limit of 1/n as n -> \infty is 0? I’m only familiar with substitution as it pertains to integrals and I know in those cases u need to change the limits. Are the rules different here?

@decare696 2 года назад

@@ezranwobodo4912 We're concerned with n->0+, not n->∞, so u->∞ is the correct substitution

@sayedsaba9738 2 года назад

@@ezranwobodo4912 you are right, you can basically make a simple argument like assume log(n)/n < 1, then n > log (n) --> e^n > n which is true for all n in N

@hach1koko 2 года назад

@@sayedsaba9738 not sure I follow your logic. I don't see how log(n)

@sharpnova2 2 года назад

awesome editing skills. i love how the values you were pointing to at the start were totally appearing on the screen

@PapaFlammy69 2 года назад

It seems like I hid my video layers in premiere pro and didn't make them visible again before rendering :(

@sharpnova2 2 года назад

@@PapaFlammy69 luckily i can mentally compute roots faster than shakuntala devi I'm a fucking calculatory genius bro

@eliaspelo8844 2 года назад

Its pretty funny when Papa is pointing no where and refering to numbers we cant see :D

@johnvonhorn2942 2 года назад

@@eliaspelo8844 Has the flame gone out?

@bowser498 2 года назад

In my Who Wants to be a Mathematician qualifier in 2020, we had the following question: which of the following is largest? sqrt(2), cbrt(3), fourth root of 4, fifth root of 5. I'm so glad I went with the second option :)

@0xVikas 2 года назад

Any number when divided by 6 can leave a remainder of 0 to 5. Out of those, its easy to see that only those numbers which have a remainder of 1 or 5 can be primes (>5). So, if p = 6q + 1 for some quotient q, then p + 2 has to be 6q + 3, which is not a prime. Hence, p = 6q + 5 (or) p = 5 mod 6 PS: Its the solution to the question shown at the beginning of the video :p

@tomkerruish2982 2 года назад

Remember the condition p>5, since otherwise p=2 and p=3 are possibilities.

@denny141196 2 года назад

@@tomkerruish2982 it's P and P+2, so 2 and 3 don't qualify

@tomkerruish2982 2 года назад

@@denny141196 I referring to the portion of the proof stating that primes can only be congruent to 1 or 5 modulo 6. Also, 3 and 5 are twin primes which would be counterexamples if we didn't include the restriction p>5.

@0xVikas 2 года назад

@@tomkerruish2982 the given restriction is exactly why my proof works, so all good :)

@mr.cheese5697 2 года назад

0:05 i have a proof for that, but it's too big for RU-vid comment, so i leave it as an exercise to reader. There really is one. Useful hint: try thinking intuitive (visual) way to figure out Euclid's proof of infinite number of primes

@MarceloGondaStangler 2 года назад

Mkkkkkkkkkk

@sachindamani 2 года назад

After seeing your comments 1st part Fermat be like, "That's my boy!"

@ricardoparada5375 2 года назад

I’ve really come to appreciate how often analysis-esque problems problems show up all over math

@PapaFlammy69 2 года назад

yup! :)

@Stixch7 2 года назад

Papa finally remembered is youtube password

@toyfabrik2993 2 года назад

Being from Germany, i can assure you that to me this sounds like perfect, accent-free English! ^^

@haakoflo Год назад

For 0 < n < e, the derivative is positive, for e < n < infinity it is negative. I'm not sure why we need to calculate the actual limits at 0 and infinity, the maximum value should be at n=e.

@lamby1907 2 года назад

I’m almost done with my 6-week college calc 1 class and I could actually partially understand what you’re saying now it’s amazing. One of my goals is to finally understand all of your videos.

@williamryman608 2 года назад

Fun fact, x = y^y^y^y^y^y^... is the same as y = x^{1/x}. I don't know what you will do with this information. I just think it's kind of cool.

@plazmyx5998 2 года назад

wait rly wow

@MrCharkteeth 2 года назад

Thank you for inspiring me to learn more about math. I love your approach in asking and exploring odd questions.

@bariumselenided5152 2 года назад

Oh, I was playing around in Desmos with this a while back and tried to answer it. I ended up educatedly guessing it was e, but I didn’t know how to prove it. It’s cool to see my question in a video, I wait with bated breath to see what you’ve got to say about it Edit: of course you need differentiation to prove it, you always seem to need differentiation to do any cool stuff in math. I wish I would’ve gotten to take calculus in high school, but at least I’m finally in a calc 1 class in college. I won’t be left behind in these videos much longer!

@mastershooter64 2 года назад

you can always teach yourself the math you need in your spare time, like you want multivariable calc? just pick up a book and finish it, same goes for any other field of math from elementary arithmetic to complex differential geometry

@anachromium 2 года назад

Don't know if it's flawed as I only do math problems for fun, but for the limit n->0, couldn't one rewrite n=1/m with m going to infinity? Then the equation would be (1/m)^m, which clearly tends to 0, as 1/m already goes to 0. The "^m" part then only speeds up that process.

@h_3795 2 года назад

Proof: p=6q+r for 0 3|p+2, contradiction if r=2, p=6q+2 => p=2(3q+1) =>2|p, contradiction if r=3, p=6q+3 => 3|p, contradiction if r=4, p=6q+4 => p=2(3q+2) => 2|p, contradiction therefore r=5

@estebanguerrero682 2 года назад

Man, your videos always gives me inspiration and push me towards studying and be free with maths, thank u so much for the inspiration

@Kualinar 2 года назад

Usually, «n» stands for an integer. If it's supposed to be a real number, then, it usually «x». Intuitively, at infinity, it converge toward 1. Thanks for the demonstration why it's so.

@jksmusicstudio1439 2 года назад

L' Hopital doesnt apply for lim(x to 0+) log(x)/x cause it's minus infinity by 0, not 0 by 0 or inf by inf. However it is known that inf by 0 is inf. A formal proof for this specific limit: Let R > e. Let d = 1/R. For all 01/R>x thus exp(-Rx)>x. Take ln on both sides gives -Rx>ln(x) or -R>ln(x)/x. This proves that for every R>e, there exists d>0 such that 0

@CAG2 2 года назад

11:58 I think you can solve it like this: we have: 0 0+ e^(log n) = lim n->0 n = 0 by the squeeze theorem: => lim n->=0+ e^(1/n log n) = 0

@PlutoTheSecond 2 года назад

Wait, if n < 1 then 1/n > 1 and thus 1/n log n > log n and thus e^(1/n log n) > e^(log n)...

@PlutoTheSecond 2 года назад

Never mind, I am wrong. log n < 0 so your method works.

@brianrobinson4825 2 года назад

This takes me back about 40 years when I analysed this function for fun, whilst doing my A Level maths. Great to see it again. I wonder if anyone could define it for negative n. Is it just discreet values or could it be graphed through the complex plane?

@Kualinar 2 года назад

In the negative direction, it can ONLY be graphed in the complex plane. The only places where it intersect the real axis is for the odd integer roots.

@brianrobinson4825 2 года назад

@@Kualinar thank you. That's what I kind of remembered.

@sleepyhaad8328 2 года назад

i recently completed my A level maths and Further Maths, and these two courses have made me fall in love with Math so many times

@WahranRai 2 года назад

We could say that is equivalent to the biggest of log(n^1/n) ---> max of y = log(n) / n ---> y' = 1/n^2 - log(n)/ n^2 y'= (1/n^2) * (1-log(n)) ---> y'=0 ---> log(n) =1 ---> n= e

@matanou7676 2 года назад

7:00 you killed me xD "who would have thought that e had something to do in here"

@juanjoserodriguezvera2405 2 года назад

When he said form and square root of Two is increasing then from cubic root of three and on is decreasing I had the sensation e was the biggest. Thank you ppa flammy a step closer to be a mwthematicin

@vsl5455 2 года назад

I litteraly just finished Analysis 1 today (litteraly wrote the exam 6 hours ago) and I find myself pleased that I was able to predict the results of this problem :)

@mehdimarashi1736 2 года назад

For the limit of n^(1/n) as n approaches +zero, note that this value is always positive and for all values of n in (0,1), it is less than n. So, the limit is sandwiched between zero and n as n goes to zero. So, it's equal to zero

@genessab 2 года назад

2 min into the video I jumped to solve this computationally (I’m a physicist how did u know) and got sucked into a beautiful wormhole of lambert W function and Euler’s beautiful proofs and extensions to the complex plane which form beautiful fractals- So much fun from a simple problem! Loved the analytical solution

@minekuchi 2 года назад

why doesn't l'hopital actually apply here?

@CAG2 2 года назад

for n -> 0+ it doesn't apply because direct substitution gives -infinity / 0. l'hopital's rule only applies when the limit evaluates to an indeterminate form like 0/0 or infinity/infinity so l'hopital's actually applies when n->infinity in this case (assuming the limit you're talking about is log(n) / n)

@cameronspalding9792 2 года назад

x^(1/x)=exp(ln(x)/x), x^x is stationary iff ln(x)/x is stationary d/dx(ln(x)/x) = (1/(x^2))*((1/x)*x-ln(x)*1), which vanishes when 1-ln(x)=0 iff x=e x^(1/x) is increasing for xe, hence it must me maximised when x=e, this means that when n is an integer then n^(1/n) attains it’s maximum when n=2 or n=3 2^(1/2) = sqrt(2), 3^(1/3) = cbrt(3) (sqrt(2))^3 = sqrt(8) < sqrt(9) = 3 = (cbrt(3)^3, thus sqrt(2)

@SebastianMageeR 2 года назад

To get the correct limit as n->0 it is easy to solve it if you use the power expansion: -log (1-x) = x + x^2/2 + x^3/3 + ... I wont post the whole prove but it is pretty much straight fordward

@Kapomafioso 2 года назад

For 0 < n < 1, since n < 1, then 1/n > 1 and so n^(1/n) can be bounded as 0 < n^(1/n) < n (replacing exponent that's larger than 1 by exactly 1 only makes the whole expression larger) According to the squeeze thm, the lim n->0^+ of n^(1/n) must be 0.

@mikejackson19828 2 года назад

"Because of the power rule, or whatever the fuck you call it!" 🤣🤣

@fedem8229 2 года назад

This is the argument I came up with for the limit as n→0 of n^(1/n). With the substitution k=1/n this becomes lim k→∞ of (1/k)^k. Now for all 1>ε>0 (we don't really care about great values of ε, so just assume is less than 1 to avoid any complications) Exists M=max{-logε,3} Such that if k>M Then k*logk>k>-logε It follows that -k*logk

@romanh219 2 года назад

Do you want to salve kind of the same problem, but in complex numbers?

@jessiechua3194 2 года назад

For the limit of x^(1/x) = e^(ln(x)/x) as x tends to zero from the postive side, we can use the substitution x=e^u, so equivalently we are evaluating the limit of e^(u/e^u) as u tends to negative infinity, which is simply 1 (since u tends to negative infinity and 1/e^u tends to infinity, u/e^u tends to negative infinity).

@jakobj55 2 года назад

6:45 dont you have to add lnx on both sides not 1?

@matejatadic1274 2 года назад

I have a question, when you were calculating the solution as n approaches 0+, isnt infinity/0+ indeterminate?

@roberttelarket4934 2 года назад

Great question! Never thought of it.

@Jj-gi1sg 2 года назад

for the limit i think I have a "better" approach.L= lim(logn/n)=lim(logn*1/n)=+inf(-inf)=-inf because limlogn=-inf and lim1/n=+inf as n goes to 0 plus. The limit is e^L=0

@gunhasirac 2 года назад

this is totally valid. I was quite confused why he says “if we use L’Hopital …” it’s neither 0/0 nor ∞/∞.

@CraftIP 2 года назад

8:49 I can't understand why you can't approach 0 from the left; negative arguments are only a problem when it's a 2n-th root

@averyhui6759 2 года назад

Log(n)/n limit can be obtained by using L'Hôpital's rule. The ratio of the derivatives is -1/n^3, which is -∞ when n approach 0+

@nafrost2787 2 года назад

You can't do L'hospital's rule, you need for it that both functions either appraoch plus or minus infinity, or zero and here, onw approaches infinity and one approaches zero.

@averyhui6759 2 года назад

@@nafrost2787 log(n) and n are both approaches ∞ when n approaches ∞

@nafrost2787 2 года назад

You wrote in your comment that you're dealing with the case where n-> 0+.

@averyhui6759 2 года назад

@@nafrost2787 Oh, you're right. I forget that and thought the problem is for ∞. Thanks for your correction.

@user-gf6nj1lh6i 2 года назад

My friend what is the music to your intro?

@chazzbunn7811 Год назад

You actually don't need to take the second derivative to determine if e is a maximum, you just need to test between your critical numbers (1, e, and infinity) using the first derivative to find where the function is increasing and decreasing. If it's increasing between on [1, e) and decreasing on (e, infinity) it's a maximum. Also, for the limit of ln(n)/n as n->0^+ recall that ln(n) is negative for positive n < 1, that alone tells us the limit is negative.

@schweinmachtbree1013 2 года назад

The limit is just a special case of the result that if f(x) → ±∞ as x → a and g(x) → ±∞ as x → a then f(x)g(x) → ±∞ as x → a where the ± signs are chosen appropriately - here a=0, f(x) = log(x), and g(x) = 1/x. (I guess we actually we need the corresponding result for one-sided limits but potayto patato)

@dan-us6nk 2 года назад

11:55 is it lazy or true to say it's undefined, 0^(1/0) cuz you divide by zero there?

@aloysiuskurnia7643 2 года назад

"no surprise that e is here" well if you are playing with exponentials, you're just *inviting* e to party

@gtxhunter9980 2 года назад

no way thats his real voice that makes it so much better!

@Assault_Butter_Knife 2 года назад

Yo papa flammy, any tips on how to keep my sanity while learning abstract algebra? Because I'm fairly sure I should be taking some kind of pills by now

@zxwy37 2 года назад

For the 0th root of 0 limit, why not rewrite the limit as lim n->inf of (1/n)^n?

@sleepysundaymorning5034 2 года назад

Without watching the video: e root of e. Explanation: we need to find the local maximum of the f(x)=x root of x for that we differentiate f(x) and set it equal to 0 f'(x) = -x^(-2 + 1/x) * (-1 + log(x)) = 0 since x is real x^(1/(x-1)) * (log(x) - 1) = 0 since x is positive log(x) - 1 must = 0 log(x) = 0 x = e is a local maximum point therefore for any positive m, e-m root of e-m < e root of e < e+m root of e+m Let me know if im correct!

@luisbelgois2102 2 года назад

You didn't check it was a maximum, but you were right about the number

@sleepysundaymorning5034 2 года назад

@@luisbelgois2102 can you elaborate please?

@andersyu4464 2 года назад

@@sleepysundaymorning5034 the derivative being 0 could mean either a local maximum, local minimum or a stationary inflection, and to make sure it is a local maximum and not the other options you need to check the second derivative

@sleepysundaymorning5034 2 года назад

@@andersyu4464 oh yes absolutely, strictly speaking you're correct, but we're given as a premise that the function is increasing until some point and decreasing thereafter, what we need to find is a local maximum in a given small range, maybe I'd be better off explicitly stating that, but otherwise you're absolutely right. Thank you!

@guest_informant 2 года назад

That's exactly how I did it. you saved me some typing. Thanks. There was an exam question a couple of years ago about differentiating x^x , so I borrowed the approach to that.

@iilunar79ii77 2 года назад

Could you use Taylor polynomials to find the limit of ln(x)/x?

@iilunar79ii77 2 года назад

You can use the first few terms and divide by x, then use l’hop to prove that it goes to -inf

@jessiechua3194 2 года назад

It's actually not that difficult/tedious to use the second derivative test and prove that y=f(x)=x^(1/x) achieves its maximum at x=e. We already have that dy/dx=x^(1/x) * 1/x^2 * ln(e/x), which achieves its zero at x=e. Then, using the product rule, f"(x)= 1/x^2 * ln(e/x)* dy/dx + x^(1/x) * d/dx (1/x^2 * ln(e/x)) = x^(1/x) * (1/x^2)^2 * (ln(e/x))^2 - x^(1/x) * 1/x^3 - x^(1/x) * 2/x^3 ln(e/x). Substituting x=e, we have that f"(e)= -e^(1/e) * 1/e^3 = -e^(1/e -3) < 0, which does indeed imply that e^(1/e) is the maximum value of y.

@spicybaguette7706 2 года назад

The limit when n approaches zero from the positive side is exp(log(n)/n), which is the same as exp(-log(1/n)*1/n). Since n approaches zero from the positive side, 1/n approaches positive infinity, and so does log(1/n), so log(1/n) * 1/n must be positive, and therefore -log(1/n) * 1/n must be negative

@raen984 2 года назад

6:30 Even if n = 0, you could multiply. You only cant divide by 0, which doesn’t apply in this case.

@raulzuniga2371 2 года назад

But in this case it would be kind of multiplying by 0 in both sides, which would kind of invalidate the equation and that's why he made the disclaimer, isn't it? Sorry if I said a blasphemy, math is not my main field.

@createyourownfuture5410 2 года назад

But he cancelled the n², which does require division by 0.

@raen984 2 года назад

@@createyourownfuture5410 He didn’t cancel n^2, he canceled 1/n^2. It’s in the denominator, so he multiplied by n^2, he didn’t divide.

@createyourownfuture5410 2 года назад

@@raen984 If I give you (1/7)*7, won't you be dividing by 7?

@esupton783 2 года назад

@@createyourownfuture5410 if you have n^2 in the denominator, it's already implied that 0 is out of the domain, so multiplying by n^2 does nothing special. If anything it re-includes 0 into the domain

@sodiboo 2 года назад

when you're reading out the first few roots at about 1:10 you point to the center and move out the way as if it's supposed to show the value, but it's just blank?? and you don't even say the values other than sqrt2 except them changing by some value?

@st_s3lios860 2 года назад

For f(x)/g(x) where f(x)->infinity in a and g(x)-> 0 in ayou can say f(x)>M near of a so f(x)/g(x)>M/g(x) near of a, and the limit of M/g(x) is infinity, so limite of f(x)/g(x) is also infinity (by comparison) Of course, you can do the same for -infinity

@sabotagedgamerz 2 года назад

I've been studying this problem and its limits and maxima and I have found negative inputs of n that result in real outputs that tend to infinity. If n = -1/x where x is a positive integer, then the nth root of n will be real. If x is an even number, then nth root of n is positive and equals to x^x. If x is an odd number, then the nth root of n is negative and equal to -x^x. This is because when we do the substitution n = -1/x, we get the result that The nth root of n = (-1)^x * x^x.

@violet_broregarde 2 года назад

Good video, thanks for sharing. I don't think we needed to check the second derivative to know it was a maximum. We knew it increased from 2 to 3 and decreased from 3 to 4. So a 0 derivative on the interval (2,4) means a peak, not a valley.

@Harkmagic 2 года назад

This looks like it would be fun to examine for complex numbers.

@Ryan-gq2ji 2 года назад

Papa, can you explain how to use the binomial theorem to prove the limit?

@PapaFlammy69 2 года назад

I can make a separate video on that, sure! =)

@mastershooter64 2 года назад

this is insane we need to show this to elon wuck

@covariance5446 Год назад

I didnt actually watch the video (yet). Just popped in to say that I just saw the thumbnail and was curious. but I was also lazy so I insta plotted it on desmos to see the shape (didn't look at the answer, though) and was curious again. Then I solved it on paper using logarithmic differentiation - and was quite delighted with the answer! Then (spoiler) I remembered what the deal is with e and it being the constant associated with maximizing compound interest (e.g., 50% twice in a year? 33% thrice in a year? etc.) and everything clicked and made sense and I felt both enlightened and stupid at the same time for not seeing it earlier. What a super fun exercise! :)

@skit_inventor 2 года назад

Couldn't you just bound log(n)/n < log(n) when n < 1 for the log limit?

@Bourbon_Biscuit 2 года назад

Fun fact:- Wuck is rated 6/9 so you know 69. Perfect rating

@thewitchking2556 2 года назад

So the reason we can’t use lhopitals is because the form technically isn’t really indeterminant it’s pretty clear top is - infinity and bottom is zero plus and that just makes a bigger negative infinity which isn’t indeterminate

@bluerendar2194 2 года назад

For one option, it should be fairly easy to do an epsilon-delta proof of the limit at 12:00. Simply bound epsilon to 1/e or less and ln(n)0+}1/x=infinity.

@nafrost2787 2 года назад

For the lower bound, why did use an argument with limits, instead of simply substitute n=1. Or even for both cases checking the sign of the derivative? We know Log(x) is an increasing functions, and the rest of the relevant functions are 1 and functions which are always positive. So it's not that hard to verify the sign of the derivative.

@hbarudi 2 года назад

what about y=x root of x and find the local maximum? will that work? Another way is to write y=x^(1/x). Good to see that with a little simple (calculus 1 level) math it was possible to find the value e for the local maximum.

@dneary 2 года назад

For f(x) = x^{1/x}, then if you take g(x)=ln(f(x)) = ln(x)/x you can find its max value easily, and since ln() is monotonic increasing, the max of g(x) corresponds top the max of f(x). This makes lim(x->0+)g(x) = -\infty easy to calculate also (as you did in the video).

@nafrost2787 2 года назад

I don't understand why the log limit was so complicated for you. We have a function which approaches zero from above in the denominator meaning the fraction approaches infinity, times a function which approaches negative infinity, so the product approaches negative infinity. And we know that 1 divided by a function which approaches zero from above, approaches infinity, because it's an easy proof from epsilon delta definition of limit.

@desertrainfrog1691 2 года назад

"---🧐 because of the power rule... Or whatever the phucc you call it." ~PapFlammy 2022

@OmgEinfachNurOmg 2 года назад

Please do this for negative or complex numbers

@ameer.a_r 2 года назад

"because of the power rule or whatever the fuck you call it" has to be the quote in the video.

@eipiwau 2 года назад

ln x is smaller than or equal to x-1 hence (ln x)/x is smaller than or equal to 1-1/x for all x>0. Taking the limit x to 0+ yields the result.

@schweinmachtbree1013 2 года назад

very clever!

@martinlehn6590 2 года назад

At 14:40 you claim that e-root(e) = e, which is not true. It is between 1.44 and 1.45.

@Gameboygenius 2 года назад

Well, seems like he marked e on the wrong axis!

@PapaFlammy69 2 года назад

oh, bummer, obviously you're right!

@tcfh2003 2 года назад

I think for the log(n)/n limit, the reason L'Hospital doesn't work is because L'Hospital is only for the cases 0/0 or "something"/infinity . Also, afaik, there are 4 conditions that must be satisfied for l'hospital to apply, but for continuous functions like log(x) and x, those shouldn't be a problem.

@tcfh2003 2 года назад

Also, by use of epsilon-delta proofs, you can show that certain functions are "faster" than others, meaning that for an undetermined result, the "faster" function dictates the value of the entire limit as the value of its own limit. It goes something like this: Log

@Alguem387 2 года назад

Betwen 2 and 4 bet is some trancendatal number like pi or e

@koenth2359 2 года назад

I Set x=1/n and looked for minimum of x^x. this more easily gives ln x=-1 so x=1/e and hence n=e

@aaronvanberkum2406 2 года назад

isn't it e^ln(f(x))=f(x) instead of e^log(f(x))=f(x) if I'm correct?

@PapaFlammy69 2 года назад

log is ln

@Vertraic 2 года назад

@@PapaFlammy69 Huh... Never seen that before. Every math class I ever took up through Differential equations, as well as my physics and circuits courses assumed log without a specified base defaulted to log base 10... ln is always log base e, while log is log base 10 Is that a research/professional difference, or is it a regional one?

@hasanansari3699 Год назад

You are so smart man

@HAL-oj4jb 2 года назад

man, 16k views in only a day, congrats on the successful upload Jens, and the video really deserves all those views ^^ Btw with your game, will wuck have more hard levels? the levels in your videos seem kind of easy to me, and I think your viewers will definitely want some really tough levels :)

@PapaFlammy69 2 года назад

thx Felix

@prebenolsen9950 2 года назад

isnt the problem of log(n)/n pretty simple to solve? just differentiate log(n) and 1/n and see which of the functions have the steepest slope when you get close to 0. You will find the slope of log(n) goes towards (negative) infinity faster, rendering 1/n "powerless". Then as the vid say you end up with e^(-inf)=0. Am i overlooking something or did my brain work for once?

@no-bk4zx 2 года назад

i love how this channel is slowly becoming a german version of dani

@matanou7676 2 года назад

why don't you calculate the limit of the derivative ?

@DeJay7 2 года назад

4:51 had me wheezing for a fair bit

@jedraszektv 2 года назад

I can tell you've been working out by those massive guns you got Jens

@PapaFlammy69 2 года назад

No lol

@tunafllsh 2 года назад

You can always use the epsilon-delta definition of limits to prove log(n)/n -> -inf

@maikopskoy 2 года назад

Yeah, lets not mention epsilon delta

@tunafllsh 2 года назад

@@maikopskoy but that's actually the most useful when you're dealing with higher dimensions and topology

@MathIguess 2 года назад

Just for some casual fun, right? :D

@roberttelarket4934 2 года назад

I did 2 days ago from Quora a telescoping limit of x^(1/x). I excluded x = 0 but included x = -1 and x > 0. The hell with 0^(1/0)!

@15hanslapesora63 2 года назад

where are we going to apply this to our daily living?

@PapaFlammy69 2 года назад

@jusinocasino11 2 года назад

Isn't natural log written as: ln ??

@Joseph125 2 года назад

Without watching the video: It's between 2 and 4, because these are both sqrt(2) Powers are involved It's probably e Edit: called it

@jakobj55 2 года назад

My approach for the limit: Take lim x-> 0+ of x/lnx. That is 0 and approaches zero from the negative. lnx/x=1/(x/lnx) lim 1/0- = -∞ = lim x-> 0+ of lnx/x MAFS (*insert meme man*)

@JacksonBockus 2 года назад

Okay I sat down and did the math before watching this video and my answer is e. Leaving it here for better or worse as I watch.

@andycopeland7051 2 года назад

Sir, this is a Wendy's

@PeeterJoot 2 года назад

Love the precise mathematics nomenclature: "... or whatever the Fkk you call it"

@anon6514 2 года назад

Can't believe I'm only now solving this one... A pleasing result; I wish I could say I was surprised, but this problem had e written all over it.

@aceleraupmimster 2 года назад

tbf i just drew the graph then used limits to show it approaches e

@Robert-el2mf 2 года назад

L’hopital doesnt apply here papa it only applies for 0/0 or inf/inf But yeah you could indeed say imo that as n->0+ 1/n is infinty and log n is -inf so its -infinity without hesitation

@PapaFlammy69 2 года назад

nah, it applies for more cases than those two^^

@farrankhawaja9856 2 года назад

@@PapaFlammy69 Wait really?

@stewartzayat7526 2 года назад

@@farrankhawaja9856 I believe it applies to 0/0 and anything/inf. Or maybe not, I'm not even sure anymore

@Robert-el2mf 2 года назад

@@PapaFlammy69 it applies for any indeterminant form as long as you simplify it and get it to its standard form which is inf/inf or 0/0 like 0•inf or inf-inf,however 0/inf or inf/0 isn’t indeterminant, for 0/inf its 0 • 1/inf which is 0•0 Inf/0 its inf • 1/0 which is inf • inf.