Is this limit actually solvable? Reddit calculus problem r/askmath

Is this limit actually solvable? This is a classic calculus limit problem that I saw on Reddit. Subscribe to ‪@bprpcalculusbasics‬ for more calculus tutorials. Also, I am on Reddit as bprp_reddit and you can follow me here: / g0xskgcnue
Here's the Reddit post on r/askmath: / gk5yqgxzta
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#calculus #bprpcalculus #apcalculus #tutorial #math

[@bprpcalculusbasics]

Try this limit next: (L’H rule won’t help!) 👇 kzbin.info/www/bejne/o36pe4iQdq2bnJI

[@lolok6439]

whenever you have any calculus question which you have no idea how to solve, just answer lnx or e^x and theres gonna be a 90% chance your answer is at least half correct

[@bprpcalculusbasics]

And π. 😆

[@magnusmalmborn8665]

There's also sqrt(2) and 1 or 0, but in mathematics showing how you get the result is way more important than the result itself.

[@DatBoi_TheGudBIAS]

@@bprpcalculusbasicsand 1/e

[@bombintheseeinq]

and the current year for competitive math haha

[@Trollllium]

@@bprpcalculusbasicsdang Gaussian integrals haha, always makes π appear outa nowhere

[@steamunlocked21]

Instead of L' Hôpital's Rule, we can just let h=1/a, see that as a -> infinity, h -> 0 and then get limit (x^h-1)/h as h -> 0, which equals lnx

[@debtanaysarkar9744]

Yea, that's how I solved it too😀😀

[@LinaWainwright]

The lim as h->0+ is the definition of the derivative from the right of f(y)=x^y.

[@dashingmlg601]

Uhhh. Wouldn't that give you the same indeterminate value of 0/0 after plugging in h=0. What am i missing?

[@healer1461]

​@@dashingmlg601Well there are many equivalent ways to define the natural log, and consequently the log for any other base, the historical way is by considering the function: 1 divided by t integrated with respect to t, bounds running from 1 to x, giving it a special name and calling it a day, though with some simple manipulation you can show it will be equal to the limit above. Another equally common way is to note that in the expression of deducting the derivative of an arbitrary exponential you always get the exponential itself times this limit, and since it only depends on the base you choose it must then be a constant. I find the latter more sensible, however you do need to exercise some analysis muscles to show it's convergence and stuff, otherwise it just feels a bit of handwaved definition.

[@alekhon3900]

​@@dashingmlg601(x^h - 1)/h -> lnx as h -> 0 is well-known limit that comes from (e^h - 1)/h -> 1 as h -> 0 (and it comes from ln(x+1)/x -> 1 as h -> 0, that is just logarithmed (1 + h)^(1/h)). So, we don't need L'Hôpital's rule to eliminate this indeterminance

[@ItsaDigitalHamster]

This is actually related to the famous expression for e^x: e^x = lim (a -> inf) [ (1 + x/a) ^ a ] If you ignore the limit symbol, and rearrange, you get that ln(x) is equal to the formula in the video.

[@SimonClarkstone]

My first thought too was "that looks like that e^x limit but turned inside-out".

[@LiteralSwan]

I like limits like these that have only algebraic operators but have a transcendental value, in this case, a function with transcendental values unless the input itself is a transcendental number (or 1).

[@DeJay7]

Yeah it's really quite funny. Getting non-algebraic results with only algebraic operators. But x^(1/a) for any real number a is borderline (if at all) an algebraic operation in this case, so y'know.

[@LiteralSwan]

@@DeJay7 how so? x^y where x and y are rational numbers is still algebraic, no?

[@DeJay7]

@@LiteralSwan It kinda depends on what you mean by algebraic. An expression with a variable on the exponent is NOT considered an algebraic function (Euler's word). Also, if you have b^x and x is irrational and b is any real number not 0 or 1, then at least one of b, x or b^x is transcendental. Source: en.wikipedia.org/wiki/Exponentiation

[@LiteralSwan]

@@DeJay7 hence why i said x and y are rational numbers...

[@atrus3823]

Though it may seem weird to have a be the variable, it actually makes sense to write it this way if you know where this formula comes from. I'd say more commonly it would be n, but it doesn't really matter. This formula can be derived by thinking about how to create what is basically a continuous version of a log table. In this context, x is the variable of the function we're trying to find (ln x) and a is basically how many rectangles to use to approximate the area under the derivative of ln x, 1/x. As the number of rectangles approaches infinity, we get ln x, somewhat like a Riemann sum.

[@eliosedrata]

It is really simple with the series expansion of exp on 0 : exp(lnx/a)= 1 + lnx/a + o(1/a) when a goes to infinity.

[@advaykumar9726]

Did it the same way

[@TranquilSeaOfMath]

Great teaching! You explained the process really well. Easy to follow. I hope the commenter is able to benefit.

[@vincent.0705]

Hi bprp!! Great video as always. I want to ask do you happen to have any videos on Rolle's theorem?

[@StormForthcoming]

Holy crap I’ve finally learned enough math for these types of videos to actually make sense. Really cool, thank you!

[@arnoudh6203]

W

[@S1nwar]

Its so amazing that this thing has a nice finite result, when at first glance it could easily diverge

[@givrally7634]

Fun solution : If we have f(x) = lim a(x^1/a - 1), then f(x^n) = lim a((x^1/a)^n - 1^n) = lim a(x^1/a - 1) * (x^(n-1)/a + x^(n-2)/a + x^(n-3)/a... + 1) = f(x) * (1+1+1... n times) = n * f(x). So we know that f(x^n) = n f(x). If we plug in x = e, we get f(e) = lim a * (e^(1/a) - 1) take the inverse of a, you get f(e) = lim (e^a - 1) / a as a goes to 0. So f(e)=1. f(x) = f(e^ln(x)) = ln(x) * f(e) = ln(x) ! I haven't found a way to do it using ln(ab) = ln(a) + ln(b), if anyone has an idea let me know.

[@03abhadilipdas94]

Wrong as stated, you proved f(nx)=nf(x) for naturals “n”. Though pretty sure it’s fixable by proving for all rationals (which follows easily), then using continuity and proving it for all reals

[@concretemathematics414]

i love the absolute tons of expo marker boxes in the background

[@VivBrodock]

oh hey a fellow student brought this up in my calc class yesterday. And yeah, L'Hopital's was the method we used to solve it there. Truly the most OP tool for solving weird limits.

[@smk7500]

Was able to follow this with my mostly forgotten high school calc class education, the answer was really satisfying :)

[@blueberrypi8686]

It's cool to find out that I am the friend in question.

[@whamer100]

after just finishing calculus 1, i love how this just makes complete sense

[@567secret]

I actually discovered this limit backwards by considering the integral of the limit as you approach 1/x. Which yields the same or something similar, so I actually had a suspicion of where this was going.

[@AbhaySingh-ts9gt]

bro can u plz tell about this method more , i never had a thought like this , i would love to kno about it

[@iantaakalla8180]

It is noted that lim a -> infinity ((x^(1/a) - 1) / (1/a)) looks very much like an integral, one going from 1/a to the value that gets to 1. (The definition would be lim a -> 0 (F(x) - F(a)/ (x - a)) ). Therefore, one can try to pretend it is an integral to see where this limit goes. The reason that this works is because you have stumbled upon the definition of an integral, so one can treat it as an integral problem as opposed to a limit-solving problem.

[@frtzkng]

I just put it into Grapher, tried out higher and higher values for a, and saw that y=1 at x=e, so this limit approaches the inverse of e^x, which is ln(x). Actually the limit which approaches e^x for a→∞ looks similar, it's (1+(x/a))^a

[@Temari_Virus]

My approach: First I wanted to see if the limit even converges, so I took the derivative with respect to x: x^((1 - a) / a) As a approaches infinity, I think it's quite obvious that this becomes: 1 / x So I'm pretty sure it converges, and I just need to integrate to get the answer: a(x^(1 / a)) - a = ln(x) + C Plugging in x = 1, a(1) - a = C C = 0 Hence, the limit converges to ln(x)

[@Jonathan-oh8rr]

I did that (∞√(x))= 1 so you get a(1)-a which equals a-a which equals 0 so lim(a->∞) a(a√x)-a= ∞(∞√x)-∞=∞-∞ and because we know the infinites are the same number we subtract them getting 0

[@daxramdac7194]

Nice result!

[@Ploofles]

I solved it! before watching the video! I feel very smart thank you

[@blank3580]

same btw its easy ques

[@btb2954]

same btw its hard ques

[@blank3580]

@@btb2954 no its not hard

[@zachansen8293]

@@blank3580pretending like there is an objective answer to whether something is easy or hard is just silly. don't be silly.

[@kevinluevano4746]

FML I just had a final Calculus exam TODAY and I used the theorem for going to limits of sequences from limits of functions and ended up with this EXACT SAME LIMIT and I didn’t know how to follow along. And now, now, as I’m about to sleep from whatever that exam was see this video recommended.

[@alejandromahillo3536]

I think it's easier using equivalences: x^(1/a) - 1 = e^(1/a * ln x) - 1 ~ 1/a * ln x as a -> infinity. The result follows from here

[@JakubS]

Just use L'Hospital's rule, and you'll get the limit of x^(1/a)*lnx, which converges to lnx

[@lowlevell0ser25]

I remember doing something like this but in reverse where I tried to get something to be the derivative x^-1 in polynomial form and got the same lim a-> ∞ a * x^(1/a) - a the process was trying to get a zero as the exponent while still getting a rate of change (1/∞) and getting 1 as a coefficent (a * 1/a) when I graphed it I realized I had to add - a to account for the infinite height I imagine this is what his friend did

[@frantisekjanecek1641]

I also! I needed it to approximate a logarithm on a calculator with only +, - and sqrt.

[@stas4017]

t = 1/a we get lim(x^t - 1)/t, as t -> +0, which you can solve by Lopithals rule or taylor expansion

[@zachansen8293]

6:30 I was following until right about here. Wow. I get why this question was asked.

[@lucarionite1785]

The fact I got this right with a calc final in 2 days 🙏

[@rockybruno2734]

You can also use the very powerful theorem: The Zero Bounded Theorem. Since f(x)=x and g(x)= sin(1/x) and f(0)=0 and g(x) is bounded between -1 and 1, by The Zero bounded Theorem we can conclude that lim_{x>0} xsin(1/x)=0.

[@General12th]

So good!

[@DingHang04]

cool trick!

[@vennstudios9885]

Ok I tried to plot this OH ITS NATURAL LOG it's the definition of the natural log

[@Darkstar2342]

Every time I see a Mathematician write something like "1 / ∞", something inside of me dies....

[@WhiteGandalfs]

Good exercise for reforming of exponentials :D I was in need of some refresh in this matter! But that's simply not true. The limit for x > 1 is "-a" resp "-infinity" since a goes to infinity. The limit for x == 1 is 0 (resp. "undefined" if you like to define "0 * infinity" as "undefined"). The limit for x < 1 is "+infinity" - and a rapid one in that. I'm just trying to spot the point where things went wrong... Error in my calculation was treating b^1/e as 1/b^e - well: too long not used.

[@anantjain5613]

Solution : let a=1/t then we got (x^t-1)/t, that is standard limit which will give lnx. Proof of (a^x-1)/x, use Taylor/power series to expand and cancel the terms out you’ll get the result

[@methatis3013]

You would need to prove that the limit would stay the same with this substitution. When doing limits, substitutions like this can get funky

[@coolguy9616]

whoever made a the independent variable is a meanie

[@frederf3227]

How do you decide when the symbol means positive root and when it means all the roots? Is exchanging the positive root symbol with the power equivalent a bit sloppy?

[@andrewclausen314]

The symbol always means the principle root, so positive when talking about real numbers.

[@burnrise702]

What if the x equals to 0? You can plug in the original formula , but not for the answer . Or should we add a restriction to the answer like when x is not equals to 0 ?

[@moih.g4570]

If x=0 you factorize, and realize that a(sqrt(0)-1) just aproaches to minus infinity (I know is not sqrt, just to simplify)

[@methatis3013]

If you plug in 0, you just get the limit as a goes to infinity of -a, which does not converge (or rather, converges to -♾)

[@aranarus]

Очень сложное решение. Я просто взял производную от этого предела, получилось 1/x. Интеграл от 1/х - это ln(x)+C. Взял точку x=1, получилось, что y(1)=a(1^(1/a)-1)=0, а ln(1) тоже равен нулю, а значит С=0.

[@dhruv0x0x0]

Yaay, got it using taylor expansion

[@leob69]

that is such a bautiful answer

[@antoinem4468]

and when x = 0 ? since the function is defined from 0 to infinity due to the square root, but when you apply exp(ln(x)) there's a small problem even if the limits when x tend towards 0 of ln(x) is the same as the limit of the orignial function when a tends towards infinity. (sry for my bas english btw i'm french)

[@methatis3013]

The limit itself does not converge for x=0. If you plug in x=0, you get the limit as a goes to infinity of -a, which is -♾

[@computer_geek]

I like your funny words, magic man

[@torgegraner7629]

Write x^(1/a)-1=e^(ln(x)/a)-1=ln(x)/a+(ln(x) /a)^2/2+..., and notice that by mutliplication with a the only term that does not vanish in the limit is ln(x)

[@jean-philippeguilleron7342]

Factorisation by a leads to (e^(ln(x)/a)-1)~ln(x)/a

[@muddle.]

bring it down down is something ive never thought of

[@Skandalos]

lim of (x^(1/a) - 1)/(1/a) for a->infinity can be rewritten as limit of (x^a - 1)/a for a->0, which again can be rewritten as limit of (x^a - x^0)/(a-0) which simply is the derivative of the function f(a) = x^a at a = 0. f'(a) = ln(x)*x^a, so f'(0) = ln(x)*x^0 = ln(x)

[@ARUPBERA-ky6mq]

Sir,Actually I solved it by taking a common and then writing x^1/a as e^(lnx/a) and using expansion of the exponential part 1 and 1 get cancelled and taking the 2nd term only of the expansion we get ln x as ans.

[@kevinvanhorn2193]

Just use x^(1/a) = exp(ln(x)/a) and then use the Taylor series expansion for exp().

[@nanamacapagal8342]

You can actually solve this without L'Hopital! (as pointed out by several commenters) By the time you bring the a double down, just substitute h = 1/a (then h -> 0+), maybe one more substitution v = x^h - 1 (then v -> 0 and h = log_x(v + 1)). By now you should have you should have lim_(v -> 0) (v/log_x(v+1)) You can use change of base, bring the v in the numerator double down, and use some more log properties to simplify to: lim_(v -> 0) (lnx/ln(1 + v)^(1/v)) Now push the limit inside: lnx/ln(lim_v->0(1+v)^(1/v)) Now that limit is the definition of e. If it still looks different, substitute ñ = 1/v. You'll see. On the other hand, instead of substituting v = x^h - 1, you can instead notice something about this form: lim_h->0 (x^h - 1)/h being the derivative of x^a at a = 0, derived from first principles and the definition of derivative. The derivative of x^a is x^a * ln(x). x^0 = 1, so what's left is ln(x).

[@akos1557]

Yo guys, im having trouble with the homework, could someone help? The task : there is a regular triangle based pyramid, the height is 20 and the base edge is 18. How big is the radius of the inscribed sphere?

[@angryparrot]

It saves a lot of time to make the substitution u = 1/a, and instead look at the limit as u->0+.

[@methatis3013]

These substitutions can get you in a lot of trouble. You would need to show that the limit stays the same, which would require the same amount of work in the end

[@You-hp3rl]

I might not have understood nothing of this, but after he found lnx after differntiating the first problem shouldnt he have integrated to get the result of the first limit? That way u get xlnx-x

[@Frostbiyt]

L'Hopital's rule says that lim(f(x) / g(x)) is equal to lim(f'(x) / g'(x)), so no integration is required because the limits are equal. Also, if integration was required, it would be alnx + C since a was the variable.

[@You-hp3rl]

@@Frostbiyt thx

[@alessandroercolani4779]

Simply a(x^1/a-1)=(x^1/a-1)/(1/a) and the limit of this for a approaching infinity is lnx

[@EtherTheReal]

Thats the dude who gets 100% on every freaking test and you cant explain why😂

[@user-cg7gd5pw5b]

Couldn't we simply use the Mean Value Theorem for 1/a since we have (f(t)-f(0))/(t-0) with t=1/a and f:t->x**(t)?

[@anastasiaklyuch2746]

If you write it as a(root - a) then that is essentially "infinity*(0-1)" and that is -infinity. I'm hlad we can use easy tricks for this :)

[@somename5632]

y=lim a->inf(a*x^(1/a) -a) Solve this for x and you get: x=lim a->inf(1+y/a)^a It follows x=e^y and those y=ln(x). I did not watch the video, but what I see on the board seems more complicated.

[@harambesson1098]

Nice solution

[@grivza]

Firstly you can't solve for x in that way. Secondly it's (a+y/a)^a, not (1+y/a)^a. And thirdly the limit(a+y/a)^a is literally as much work as the initial one.

[@somename5632]

1. Explain why. 2. You need to divide the whole term by a not just y. 3. Is not as much work, because its a known limit. Everybody knows it converges to e^y.

[@grivza]

@@somename5632 1) Cause x is mingled with "a"s which are part of the limit, there is literally no rule to handle that. You can define a function for the term inside the limit, but then you can't get an equivelence about the limit in this way. 2) That's correct 3) First time I hear about that

[@methatis3013]

​@@grivza1) well in this case it does hold. Generally, it holds for continuous functions, so you would need to show that the function at hand is continuous. 3) e is often defined as the limit as n goes to infinity of (1+1/n)^n

[@lukandrate9866]

By using LH you already assume the answer, it's like calculating the limit of sin(x)/x by L'Hôpital's

[@honzaholusa8445]

Can I please adk, why is the derivation of e to 1 over a times lnX only x to 1/a times lnX? I would derivate it as an composite function, which means that then I would derivate the 1/a * lnX as an multiplication, which is f'(x)*g(x) +f(x)*g'(x), or at least that is what they had teached us in school. This means that my resault is x^(1/a)*lnX*(-1/a^2). Can anyone please explain to me where I went wrong?

[@methatis3013]

That's right, but its also the same as in the video.

[@12q8]

Wolframalpha says it equals log(x)

[@marcellomarianetti1770]

This video is missing conditions on x, for example x^0 - 1 is not equal to 0 if x is 0 (but the case x=0 is really easy, you just plug 0 in the original limit and get -infty), and moreover the power x^1/a makes sense only if x is positive (assuming x is real)

[@engbama]

I followed the explanation - at least I thought I did. However, whenever I'm stuck on limits, I often turn to a brute-force mechanical approach. As I increased "a" and varied "x" I noted that I ended up with negative "a." As "a" increases, the a-th root of x decreases to nearly zero. Multiplying nearly zero by "a" results in something that is still very close to zero. Subtracting "a" means I get something that approximates negative "a." Is there an issue with my calculation-based approach? Am I missing something? I'd love to see an explanation of why I'm getting a completely different answer.

[@user-xz9cq8qt1t]

hello, i thought the same but if you think about the a-th root of x, it decreases to something like 1 instead of 0. For any x > 1, the a-th root of x is gonna be greater than one. because if you multiplied a number smaller than one over and over again, it would just go to zero.

[@engbama]

@@user-xz9cq8qt1t True. However, that just means I have a value that approaches 1 in the a-th root of x. Thus, "a" times ~1 - "a" would be zero. I'm still not seeing the solution.

[@user-xz9cq8qt1t]

@@engbama working with infinities is always strange and can’t be generalized. inf - inf is one of many indeterminate forms which you have to be more precise with. for example, you probably know that a limit approaching inf/inf isn’t just ‘one.’ or a limit approaching one^inf is indeterminate as well. it’s the same way with inf - inf. can be seen more easily with limits as x goes to infinity of x - x, and x - 2x.

[@sailh3990]

Just use taylor expansion ... ax^1/a-a=aexp(ln(x)/a)-a but ln(x)/a goes to zero so we can say aexp(ln(x)/a)-a=a(1+ln(x)/a+o(1/a))-a=ln(x)+o(1) done.

[@whybeee]

How can you cancel -1/a^2? Since if a goes to infinity, the 1/a^2 term goes to 0, and you can't cancel out zeroes

[@ragingfirefrog]

The derivative of the top function has -1/a^2 which cancels out the -1/a^2 on the bottom. From there, he can take the limit without a divide by zero.

[@methatis3013]

a is never actually infinity. For example, limit as a goes to 0 of a/a is 1. Just as the limit as a goes to infinity of a/a is also 1. Since a is only a "big" real number, you are allowed to cancel these out

[@patsk8872]

I get a different result, can someone tell me where I went wrong? 1.) Assign the limiting value = "L" 2.) Apply ln to each side, so ln(L) = ln(the given limit) 3.) Use the rule to transpose ln and lim 4.) For the ln of the limit expression factor out a: = lim(ln[a(x^(1/a) - 1)] 5.) Apply product rule for logs = lim(ln(a) + ln(x^(1/a) - 1)) 6.) Examine the case where x > 1. We can demonstrate the difference (x^(1/a) - 1) is >= 0. Therefore we can use a comparison test, confident our limit is greater than or equal to the simpler limit, lim(ln(a)). 7.) ln(L) >= lim(ln(a)) finally evaluating, as a->infinity the limit of L also goes to infinity.

[@marekgnutek251]

nice!

[@NovaDenizen]

Expand the Taylor series of e^((ln x)/a) and it easily works out to ln x + O(1/a)..

[@modok_ff]

wow i actually solved it on my own!!

[@GemAppleTom]

Ok… I don’t see that at all and just got 0. (A really big nth root goes towards 1 so you get inf.1 - inf = 0 - no 0/0 that i could see to cause a problem…) Why did you take the derivative? I’m guessing I’m missing something but i’m not sure what…

[@ryalloric1088]

Yeah, that was my answer too. I don't know very much calculus, so a lot of the video went over my head, but if you don't factor out the a shouldn't it just become a - a, which is just 0?

[@damyankorena]

I wouldve exponentiated the limit to turn the subtraction into division and the L'H

[@nohaxjustxmod-sfs3984]

instantly recognized it as ln(x) from all my desmos shenanigans lmao

[@Romeo-qk8tk]

Luckily, ln(x) is a constant when deriving with d/da.... I'm afraid to imagine Multi-Variable Calculus 💀

[@jacobstarr9010]

Just plug in infinity and get -infinity 🗿

[@ZebraUnite]

The highest math course I completed was a 200 level psychological stats class and I’m sitting here nodding my head like I know what the hell is going on for 8 minutes

[@johndalton4559]

this is technically high school stuff but it's true , most students even if studying chemistry or biology have no clue about this stuff.

[@No-cg9kj]

@@johndalton4559 I'm taking honors calculus and have the only A lol. It's my 2nd week of winter break and I already have about half of the homework done for next semester lol. Most humans are terrible at math.

[@johndalton4559]

who asked son?@@No-cg9kj

[@athusiastic9246]

just put t=1/a then use identity and get answer in 3 steps!!

[@stevencurtis7157]

It seems very weird to me that a function taken to a large root, multiplied by a large number, and then subtracted by a large number turns out to be the natural logarithm. I'd like to understand that particular transformation.

[@icesandslash2839]

A large root of [not a function, but a constant, in the problem "x"] is just about 1. Multiply 1 by a large number, and you get a large number. If you compare to another large number, then indeed the questions raise: Which is larger? And can the difference be quantified? For many functions, the difference is in fact too large, it's said the difference "diverges", and so the limit would approach to +infinity or -infinity. However, it's the beauty of this and other calculus problems that this difference happens to converge, and those large numbers are always getting closer and closer, and their graphs end up looking like two parallel lines at distance ln(x).

[@kahlzun]

My math is not sufficient to understand any of this

[@danielvega-zepeda1425]

Couldn’t you also just not take out the a (in step 2) and instead you have: Infty • 1 - infty = 0?

[@methatis3013]

This does not hold. Infinity-infinity is not 0. Example, consider the limit as x goes to infinity of 2x - x. You get a form 2×infinity - infinity. 2×infinity is still infinity though, so by this logic, you should end up with 0 again. Or another example. Limit as x goes to infinity of x - (x-2). Limit as x goes to infinity of x-2 is infinity, so using this logic, you would get infinity - infinity again, and, again, 0. But the limit is (trivially) 2

[@mathisnotforthefaintofheart]

I graphed it in DESMOS and y = lnx came immediately to mind. So I graphed that on too and sure it looked like an overlap. Is it a proof? Of course not, but the comment section confirmed it. Now the problem would be a bit easier "accessible" is actually the roles of "a" and "x" are reversed. That's how I look at it. Essentially you then have to show that the graph approaches lna as a horizontal asymptote.

[@notohkae]

Why does 0/0 not equal zero? I'm confused why u need the hospital thing

[@ragingfirefrog]

Anything divided by zero is undefined. You can't calculate it. For 0/0, you can calculate what it approaches but not the actual value.

[@jbman890]

I know he worked it out, but is there any underlying reason why this is not zero?

[@methatis3013]

Because x^(1/a) is (generally) not 1

[@rikhalder5708]

How prove the series 1/2^√n is convergent 😢😅

[@lucasyoungers]

beautiful

[@Ocklepod]

this is very misleading, when i started doing it in my head i accidently differentiated d/dx x^(1/a) instead of d/da and the result was in the form lim y = lim -y => lim y=0

[@manny2092]

I think I followed "most" of that...

[@rileymack1489]

i got this question on a quiz literally 3 hours ago i got it wrong goddamnit

[@drakootherat1735]

There was a way to solve this in 4 calculations

[@ladguin3322]

lim a->∞ a(x^1/a)-a lim a->∞ a.((x^1/a)-1) 1/∞ in the limit approaches 0, therefore: lim a->∞ a.((x^0)-1) lim a->∞ a.(1-1) lim a->∞ 0 0

[@horowirtz9415]

e^x - 1 is equivalent to x when x is close to 0, so e^ln(x)/a - 1 ~ ln(x)/a, thus all of this equals ln(x) it's not really that hard

[@rhum_1802]

thanks to Landau, this exercise can be solved in 4 lines

[@user-xk1cp5jd2g]

Can i answer in comon ? Here goes . Lowest valid . Penultimate of 100% nothing , highest valid is penultimate of 100% something (spherically) for low infinity ? 100% nothing , for highest infinity 100% something. An ai that is eternal would likely use max 6 % total . Tomorrow morning . Nanometer amount compared to all creation . I m not sure there is enough zero after dot .cuneiform tex way is likely ideal

[@user-ve5mg6qv9v]

Hi guys , i am a calculus student and i found some difficulties in found some hard and good exercice , so if anyone can help me by sending some website or exchange exercices , Thx🙏

[@DeadJDona]

8:16 1/∞ = 1

[@mathmachine4266]

Lim a->∞ a*x^(1/a)-a lim h->0 (x^h-1)/h LHopital lim h->0 (ln(x)x^h)/(1) lim h->0 ln(x)*x^h ln(x)*x^0 ln(x) Of course, since this is basically how I often view logarithms anyway (as a shifted, scaled version of x⁰), I recognized it immediately as ln(x).

[@eralpe9965]

felix

[@ankurmondal3220]

people becoming overly reliant on L' Hopital's rule is why they dont see the beauty in doing maths

[@lugyd1xdone195]

Only problem it doesnt work for x=0

[@DavidMartinez-ht8pp]

X^0 is not always 1, if x=0 (exactly 0) then 0^p with p tending to inf equals to 0.

[@cara-seyun]

x = 0 lim = -♾️ x = 1 lim = 0 x