Mean Girls "The Limit Does Not Exist" Math Problem
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- Опубліковано 19 вер 2022
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Make a video on Riemann hypothesis
I'm glad you looked at the one-sided limits individually. Way too many UA-cam channels just stop after misunderstanding L'Hôpital and claiming the original limit cannot exist because the new limit doesn't exist.
For those unaware why that's wrong:
It's because that line of reasoning doesn't apply for when the new limit oscillates like (x + sin(x))/x as x tends to infinity.
The actual limit in my example is 1, but applying the other UA-camrs' logic would make it seem like that limit doesn't exist.
L'Hôpital only guarantees equality when the new limit exists. So, if the new limit doesn't exist, you have to find a different approach or break it up into two limits like what Michael Penn has done! :)
I think the actual name is de l'Hôpital
I believe you can't use L'Hopital rule if your initial limit does not go to 0/0 or infinity/infinity
@@alxlg "de" just means "of", like "William of Orange". l'Hopital is fine to use as a family name to refer to the guy :)
Can you please explain how looking into the side limits solves the problem in l'Hospital's theorem? Specifically, how does checking the side limits of the new limit prove that the original limits doesn't exist?
@@user-yv1qs7sy9d because if the new expression tends to ∞ (or -∞) then de l'Hôpital tells you that the original expression does the same. In other words, you can apply de l'Hôpital if the new expression converges to a finite limit or diverges to ∞ or diverges to -∞.
such an easy problem for the sudden death final question of a math competition
I watched this movie for the first time at a science summer program. Everyone demanded the movie be paused and solved it 😂
This is, of course, the most fetch of any maths problems
The "fast and loose" version is about 10x more rigorous than how any physicist uses Taylor expansions for this kind of problem in practice lol. Which is to only calculate lowest order for the numerator and denominator and see if the lowest powers cancel out or not. I even found writing out the Taylor series as an infinite sum incredibly jarring: I almost always think of the Taylor series as first two or three terms only.
Another way to write just the low-order terms is to write it as f(x) = a_0 + a_1 x + a_2 x^2 + O(x^3) (assuming 2nd order). Still avoiding the infinite sum notation, but this version is technically correct. Or you could just write Err(x) for the error term, that's also something I've seen done!
Shush, I thought we all had agreed to keep that a secret? The mathematicians can never know!
thats a little bit disrespectful
Every infinite series has to be written out all infinity terms to be properly checked. There is absolutely nothing rigorous about assuming anything in this world. If you havent counted to infinity, then you can not by any means whatsoever claim that there are infinitely many numbers. This is logic 101
@@pandabearguy1 i can see you are new to mathematics. Maybe pick up a book before asserting such ridiculous a statement.
Oh my god Michael, you can't just ask why limits don't exist
Ngl, I actually was able to follow along with this. Watching videos like these has definitely helped with me remembering all my calculus studies. It’s been far too long since I took calculus classes.
I'm just grateful that Michael Penn exists!
You have to demonstrate his existence and uniqueness before being able to say that!
@@aronbucca6777 His uniqueness is a sufficient, but not necessary condition.
you don't understand how happy I am that this crossover happened
That was meanly fun.
Thank you, professor!
It's been a while since I've done this kind of maths and for some reason I was never taught about the one-sided limits in this way, and I was also never taught about L'Hospital's rule. I wish I would have known it before because it seems very useful. Anyway thank you for this video, it taught me a lot!
I remember applying L'Hôpital and one-sided limits back when I took Cal 2 in college for this problem, it made me feel like a genius lol thanks for the great vid!
Dude I’m doing this kinda stuff since my eighth grade first semester calc 1 class
You can avoid being “sketchy” by adding remainder terms in little o notation to your finite limit expansions, but that’s rigor for the sake of rigor.
You’re concision is appreciated
You did exactly what I am teaching my students not to do. There exist cases where the limit of f'(x)/g'(x) does not exist while the limit of f(x)/g(x) exists. (Althought as the limit in the video is not one of them).
What the De L' Hospital Theorem says, is that (under the other conditions) if the limit of f'(x)/g'(x) exists then the limits of f(x)/g(x) exists. (But the inverse is not in general true). On the other hand De L' Hospital Theorem applies also in side limits (actually this is the method that the theorem is proved), so
in the case of the video, we may apply the De L' Hospital theorem to the side limits, and in the end arrive to the conclusion that they are different therefore the initial limit does not exist.
10:41
Good
Good
If he doesn't have this on a T-shirt he's missing a trick!
My mathematics education is sadly lacking, and though I had read about l’Hopital’s rule, I never really got it until ~1:45 of this video. Totally just clicked.
Thanks!
OMG I TOTALLY NOTICED THIS DURING QATCHING ILYSM
I played it even more fast and loose and just did the one-sided limits for the original problem. From the right, we have a very small negative number in the numerator and an even smaller positive number in the denominator, so it's negative infinity. From the left, it's a similar situation, but with a positive numerator, so positive infinity.
Having +sin _x_ instead of -sin _x_ in the numerator would've made for a more interesting problem, as the linear _x_ terms in the Taylor series would cancel.
The first time I watched Mean Girls my friend paused with this limit on screen and asked me if I could solve it. I got it wrong because I assumed the linear terms cancelled :(
I took the lazy approach since the limit can be broken up into a -csc(x) term which clearly diverges at zero, so the limit doesn't exist. Done. Maybe in certain scenarios you have to worry about additional terms, but it has been a long time since my calc II days. So I will leave it to someone else to correct me if I am wrong!
Another math channel doing this problem is wonderful.
Can finally solve a Michael Penn problem in my head 😭 persistence for the win
You are the exact spitting image of Tom Scott.
Trademarked red shirt and everything.
Nice explanation 👌
As Taylor approached Sine of X, he asked the function, " Can you be approximated because you're differentiable? Or are you differentiable because you can be approximated?" He then proceeded to use his cursed technique, "Infinite Differentiation," to try to reduce the function to zero. However the function Sine of X is cyclo-differentiable, and so by contradiction the function stated, "Stand proud, you're strong. But nah, I'd win."
Brook Taylor fell to his knees. In his dying moments, he uttered the words...
"Domain Expansion."
Because when it comes to differentiable functions, always bet on Taylor. The ones who founded calculus, the ones who formalized the derivative, they would all bear witness to this penultimate technique:
"Infinite Series Expansion!"
With this, Taylor stood up and said, "Throughout the heavens and the earth, I alone can differentiate."
-/watch?v=Ow6Y-g4nRls
Instructive, thanks.
An even sketchier way of looking at the problem from about 1:26 is to say that ln(1-x) - sinx tends toward 0 - 0 = 0, while the denominator tends toward 0 x 0 = 0² which is a "higher order" of zero, so we should expect the function to behave like k/x when x is very small. That's enough to suggest to me that the limit as x→0 cannot exist, and you can do that in your head.
Doing some kind of "hand-wavy" analysis has always served me well before proceeding to invest time in more rigorous investigation, because I'm then surprised if that turns in a different answer. A different answer would motivate me to double-check as I almost always find an error somewhere in the more complex working.
This is good intuition for why the limit DNE
What do you think about (Sin(x) -x)/x^2 as x -> 0 ? Is 0 - 0 =0 now or is it 0^3?
This is a very dangerously wrong reasoning you presented.
@@lavneetjanagal No, of course not. it should be obvious that the numerator is of the order of x^3 and the denominator is of the order of x^2. The function will look like x for very small values of x and the limit will exist. It's not that difficult (or dangerous!) to do the approximation in your head as I suggested.
To check the approximation, take the Maclaurin expansion you'll see that the first term of (sin(x) -x) is -(x^3)/6, and so the function you presented gets closer to -x/6 when x is in the neighbourhood of zero.
[Edit] I can see that I made the mistake in my original post of writing "0 - 0 = 0", when I actually meant "- 0 - 0 = 0" since ln(1-x) is negative for small positive x. Apologies for confusing you.
While this sort of reasoning usually works, there are cases where it fails - spectacularly so. I remember 1st year of university, Calculus I, my professor gave us a very specific limit which I can't recall but I can try to look it up later. My intuitive reasoning similar to yours convinced me the limit was 0, I even checked with a calculator for extremely small values of x (a tactic I used to use whenever I had a limit or summation or whatever and the worksheet didn't have the solution) and it approached 0. Time for the rigorous proof and voilà, the limit was actually 1/24, which isn't even that small of a number for a calculator to "confuse" with 0. I talked to my professor about it and in the next class he said there had been "doubts" regarding said limit, then went on to prove the limit was 1/24 and say calculators make rounding errors that sometimes yield incorrect results like in this case.
@@yvltc Yes, indeed. If my suggested "in the head" way of estimating these sort of limits always gave the right answer, there would be no need for a rigorous derivation of the limit. The point is that having an intuitive idea of what the limit ought to look like acts as a sanity check when we perform the real calculations and helps us spot when we have probably made a mistake in that calculation, so prompts us to recheck our work. I think your anecdote is a good example of that process at work.
The second method is not sketchy, that's the method we use in France instead of l'Hopital's rule. It's called "developpement limite" fr.wikipedia.org/wiki/D%C3%A9veloppement_limit%C3%A9 (it's only in French sadly). There's just some notations that are missing in Michael's example for it to be rigorous !
Yess! Also in Italy we use this method with the little o notation, we call it Peano remainder and it's perfectly rigurous and justified by theorems.
I don't think this has something to Do with the country you are from xD everyone uses this. I'd evem say that, since L'Hopital was french, you also use that rule in france ;)
@@samueljele That's what's funny ! I've never heard about l'Hopital before watching American Maths videos ahah.
Like the concluding "Good place to stop" Short videos are sensitive to attention span, retention capacity of viewers on challenging at the same time entertaining subjects. Thank you☺
Agreed. He does this for all his videos, there's even a novelty account @Good Place To Stop that points out the timestamp where he says it.
@@iooooooo1 Indeed! I've seen folks stop eating while still enjoying, Trying to saturate oneself maybe inadvisable in some situations!☺
Fast and loose works if you can demonstrate that for the expansion everything to the right of the leading terms don't affect the limit, which seems pretty doable
Minor simplification. 2 sin x cos x is equal to sin 2x. This eliminates the need to worry about the behaviour of cos x as x tends to zero.
Its a trival thing but it eases the cognitive load a tad which is alawys a good thing
so if I was in Lindsey Lohan's shoes, i would have won because I answered "The limit does not exist" but if I was asked for proof I would have failed because I would have just written "it's undefined/a singularity"
What about the limit as x goes to zero of x/x? :)
One sided limits can be calculated without L'Hospital rule and Taylor expansion (both need derivatives)
We will get the limits which appears in derivatives of ln , and trig functions such as sin , cos , tan
but we can calculathe this one sided limits without derivatives
we can apply the result sinx/x = 1 as x approaches zero directly to simplify the expression by dividing both numerator and denominator by x^2 from which the expression reduces to ln(1-x)/x^2 - 1/x;
Now by applying Taylor series expansion -(x+x2/2++x3/3+......)x2 -1/x expression reduces to (-1/2-2/x) as x approaches to zero from which it is clear that the limit is not existing.
A non-sketchy way to do it with Taylor series would be to write the first n terms of the expansion plus o(x^n), in asymptotic notation. Then, after dividing the numerator and the denominator by x^n, you'd get o(x^n)/x^n which cancels out to 0 as x approaches x0 (0 here) - by the o property, and the remaining terms would give you the limit in a usual way.
Note: f(x) = o(g(x)), x->x0 means for all e>0, there exists d>0 s.t. for all x in d-vicinity of x0, excluding x0, intersected with f's domain, we have |f(x)| < e|g(x)|.
If g is nonzero anywhere at that vicinity with x0 excluded, and has the same domain as f, then this is equivalent to lim(f(x)/g(x))=0, x->x0.
Jee student??
@@hrishikeshpatel6670 Hi! Did you mean JEE in India?
(I'm not-I'm a student from Ukraine.)
Le jee students* we eat sums li this in our breakfasts😂
Jee advance be like we didn’t even consider this as a problem
The Taylor approximations aren't cut at the same places, but second order + O(x^3) should be sufficient to get something rigorous here. In particular, x^3*convergent is O(x^3), and x*O(x^3)->0 as x->0.
As x->0, sin(x)->x (ie sin(x)/x -> 1), so the limit will be the same as for (ln(1-x) - x)/x^2. It's pretty easy to visualize that it approaces negative infinity for x->0(+) and positive infinity for x->0(-)
Him : Lopitals rule
Jee Aspirants: L Hospital🏥🚑 rule
Could you do a video of the proof of l'Hospital?
I just split it into two fractions, the right fraction is negative sign over sine squared, which reduces to negative one over sine, which as X approaches 0 becomes-1/0 which is undefined therefore the limit is undefined
I'm afraid this logic doesn't work. Just because there is an undefined expression like -1/0, doesn't mean that the limit is undefined. By that same logic the limit x/x as x approaches 0 would also be undefined, but it's 1.
@@tc14hd23 interesting point but I don't think this is relevant because this is much more like the limit as X goes to zero of one over X. In the limit you're taking there's a cancellation for all numbers that are not zero, there's no cancellation in the number flies off towards Infinity and negative Infinity
Very good explanations, nice reminders of these calculus ideas. So many videos assume you still remember all of these things
No need to use l'Hopital or Taylor or even to know derivatives. Just factorize 1/sinx and divide by x numerator and denominator : [ ln(1-x)/x - sinx/x] [ sinx/x] -> -2 (all well known elementary limits). Then we're left with 1/sinx, whose limit to 0 does not exist obviously.
This isn’t sketchy provided you use the little o(x^4) notation.
A much nicer problem would be lim (ln(1-x)+sin(x))/(1-cos(x)). The Taylor expansions yield the correct result -1.
No big issue or anything but for some reason the l'hospitals in the board was killing me lol.
Riemann hypothesis solutions
I may be seeing this incorrectly, but didn't he forget to add the negative sign when taking the derivative of ln(1-x)?
It should be 1/(1-x) × (-1)
Nevermind he factored that into the denominator lol.
If you set yourself limits in life you will forever be limited
Isn't L'Hopital rule essentially just a simplified version of Taylor in this case? You get Taylor series by taking derivatives N times. As for "fast and loose", maybe you could've used o(x^n) remainder to be slightly more accurate with it.
good video :)
Multiverse of madness
I wonder where Carolyn Kraft in the movie came up with the answer -1.
Is the first step required or can you just plug in and realize cos0 is 1 to get lhospitals
You can try problems from ISI (Indian Statistical Institute) B.Math and M.Math entrance exams ...
or even Chennai Mathematical Institute (CMI) entrance questions
they are brilliant
Bleh
The simpler way to do this is to recognize that, for small x, sin(x)->x. This makes the equation become (ln(1-x)/x^2)-(1/x). From this, its obvious that the limit is undefined.
Isn't the taylor expention and l'Hôptial rule doing basicly the same here lim u/v = lim u' *x/v' *x = lim u'/v'
If the limit after l'Hospital turns out to be nonexistent, you cannot say, that the original limit is nonexistent...
In this case it works - is it because there is a non-said default implied "I was actually doing the side limits from the beginning, I just didn't bother writing it separately"? Like, if you did it rigorously, you find the two side limits separately from the beginning using l'Hospital, they turn out -inf and inf. Realize the side limits differ, so the actual limit does not exist. Is this the case? To me it makes sense this way. Suddenly there is no problem with the l'Hospital and non-existent limit.
2:53 What is the difference between "the limit does not exist" and "the limit is infinite"?
It's L'Hopital's rule that is sketchy ... the three term O approximations simply tell us the truth !
L'hospital's rule? Lol
I guess the small angle approximation will also yeild similar results.
I guess it is the same as using the Taylor series.
( I took a antenna class in college.
It is basically the solution to the entire class.
When you can't solve a problem by hand, just use the small angle approximation.
Congratulations, you passed the class.)
Small angle approximation is a first order approximation, so whenever you need higher order expansion it won't work
@@deept3215
In this case, it will work since we are taking the limit to 0.
All the other terms will be much smaller than the 1st order term.
Obviously, it will not work if the angles are not small.
@@porcorosso4330 That's not how it works. As I said small angle approximation is a first order approximation. Now consider for example lim of (x-sin x)/x^3 as x approaches zero. Small angle approximation would say that's zero. Is that right? What about (x-sin x)/x^4 ?
@@deept3215
No, in your example we should look beyond the 1st term.
For these 2 example, it should be enough to expand to the 2nd term.
@@porcorosso4330 And that's exactly what "small angle approximation is a first order approximation" means. And why it doesn't "always" work.
Just doing this quickly in my head, but doesn’t it just approach -2/x as x → 0?
That's quite right, and IMHO it's good to get an idea in your head of what to expect if you were to do a more rigorous investigation.
One can use equivalence rules. sin x is x. ln(1-x) is -x. cos(x) is 1 - x²/2
fix misprint on cos(x)
Huh? I've never heard these called "equivalence rules". These are essentially the first terms of the Taylor expansions. And that's essentially what Michael actually _did_ do in the second part of the video.
And BTW, "cos(1-x) is 1 - x²/2" is simply wrong. You have not only one, but several errors / typos in that claim...
You're correct that for small x, sin(x) ≈ x and ln(1-x) ≈ - x, but for small x, we find that 1 - cos²(x) = sin²(x) ≈ x².
I'd like to demystify "equivalence rules". It's called the fundamental theorem of engineering. You're welcome.
@@RexxSchneider I see no contradoction here.
denominator is best turned to sin²(x) ≈ x².
I meant this.
What I meant was go as simple as possible. Not to use overkill. Like use L`Hopital`s rule or Taylor series.
Just note. One can use 1 - (1 - x²/2 )^2 as well. it will also turn to x^2.
Note misprint fix in first comment.
@@patricius6378 Equivalence rules is standard tool for limits. No mystics in it. So I don't get your point.
Use limited developments of order 1 in x and done in 1 min!
ln(1-x) = -x (could integrate the geometric series 1+x+x2+... = 1/(1-x))
sin(x) = x
1-cos(x)^2 = sin(x)^2 = x^2
Then lim = -2x/x^2 = -2/x
x->0+ => lim = -inf
x->0- => lim = +inf
When would you need to know this in real life, outside of teaching it, or taking a course on it?
I enjoyed this video, even though I have never heard of the Mean Girls.
The man's name was L'Hôpital, not L'Hospital, as your correct pronunciation suggests you know. Because the circumflex over the o originally indicated that an s had been left out, I suppose you could argue that you haven't made a mistake, but simply used an antiquated spelling that you saw in some old document somewhere.
The marquis himself spelled it L'Hospital.
@@reamick Thanks. I'd be genuinely interested to know how you know this. Have you seen manuscripts written by the marquis himself? If not, the "Remarque orthographique" on the French wikipedia page observes that at least one scholar closely associated with the marquis used the spelling with circumflex. I wouldn't put too much store on printed title pages of the period: printers had their own style conventions that they would have followed when two spellings were regarded as equally valid. Though none of this is quite as interesting as the question of why Professor Penn chose to depart from the spelling most commonly accepted by today's mathematicians.
could you say that the limit is undefined
2:24 but derivarive of ln(1-x) is -(1/(1-x))
-(1-x)=x-1
Note that he mentions that he's changing the order of subtraction when he writes it
That was so fetch.
I thought L’Hopital was taking the derivative of the top AND bottom
I'm a be honest with you, I probably would've missed turning 1 - cos^2 into sin^2
The guy looks like a very intelligent version of Forrest Gump
lim of numerator is 1, while limit of denominator is 0 and since the denominator is always positive in a nbhd of 0 (not containing 0), this implies the limit is infty.
Is this not enough?
sin x in the numerator will be positive from the right and negative from the left. Therefore, the numerator is negative from the right and positive from the left. This means the Lim as x approaches 0+ is negative infinity and the limit as x approaches 0- is +inf. This means the limit does not exist. Also, the limit of the numerator is 0 not 1.
@@thesecondderivative8967 yeh i had a bit of a brain fart and did ln(1) = 1
The limit of the numerator is 0. Check that for small x, we find ln(1-x) ≈ - x and sin(x) ≈ x. So for small x, the function looks like (-x - x) / (x²) ≈ - 2/x which has no limit at x=0 (it's both +infinity and -infinity).
I've not seen the movie so to save anyone else having to look for it its in this clip here: ua-cam.com/video/EligNcjdyyI/v-deo.html
is it a problem to conclude during L'hospital application that if lim(f'(x)/g'(x);x->a) DNE then lim(f(x)/g(x);x->a) DNE?
Remember that L'Hopital is formulated as the equality of onesided limits of f/g and f'/g'.
So if the limit from below or above of f'/g' exists in |R-bar (i.e. |R with +- infinity), then you can conclude that the same is true for f/g (from the same direction). That means if both, the limit from above and below, exist in |R-bar, but are not the same (e.g. one is +infinity and the other is -infinity, as in this example), then both the limit of f'/g' and the limit of f/g do not exist. But if the limit from above or below of f'/g' does not exist because it is for example oscillating, you can't conclude the behaviour of f/g.
I'm not 100% sure, so please someone correct me if I'm wrong.
@@samueljele i think U can use L'Hopital in this case if u cacluate the original limit from each direction, and then apply L'Hopital for each direction ie:
L⁺=Lim{f'(x)/g'(x);x→0⁺}=-∞ ⇒ Lim{f(x)/g(x);x→0⁺}=L⁺=-∞
L⁻=Lim{f'(x)/g'(x);x→0⁻}=+∞ ⇒ Lim{f(x)/g(x);x→0⁻}=L⁻=+∞
L⁺≠ L⁻⇒ Lim{f(x)/g(x);x→0} DNE
@@samueljele observe non ex.:
f=x^2⋅sin(1/x)
g=sin(x)
f'/g'=(2x⋅sin(1/x) + cos(1/x))/(cos(x))
lim(f'/g';x→0) DNE b/c lim(cos(1/x);x→0)
but as x→0, x/sin(x)→1 and thus x^2/sin(x)→0,
and since sin(1/x) is bounded between -1 and 1, we have
lim(f/g;x→0)=lim([x^2/sin(x)]⋅[ sin(1/x)];x→0)=0,,
@@orenfivel6247 yes exactly that's what I meant.
Your Taylor series for the numerator is not correct because you kept x^3 terms coming from sin(x) but you neglected same order term coming from log(1-x). Of course, this is immaterial when taking the limit, but I discourage my students from writing incoherent Taylor expansions, meaning expansions where different contributions are expanded to different order in x
Fun fact: L'Hôpital's rule wasn't actually invented by l'Hôpital but by Johann Bernoulli who then sold it to l'Hôpital
Maybe Michael could make purchasing Fermat's Last Theorem from Andrew Wiles a tier on his patreon
Or just (-x+o(x)-x+o(x))/(x+o(x))² = (-2x+o(x))/x²= -2/x (thus the limit at x →0 doesn't exist in ℝ ∪ {-∞, +∞} with the usual topology).
I have no idea what any of this means
Be sure to note that the reason this works is because the one sided limits have the form #/0, meaning an infinitely small pos number is basically inf and an infinitely neg number is neg inf. Otherwise, this logic would imply that a straight line has a DNE limit everywhere.
the only tricky thing in this problem is the proof of the L'Hospital rule :)
Jee aspirants be like are you comedy me 🤣🤣🤣🤣
Why am I getting -1/2 after using L'hopitals rule
Can't you just consider the limit of 1/sin, which blows up the whole thing?
because lim (A + B) = limA + limB if and only if limA and limB exists. Because the limit of 1/sin doesn't exist, you can't just isolate that term.
1:24 wait... the world didn't blow up?
I wonder when we analyze the limits, why we cannot just say the limit can be EITHER a positive or a negative infinite value. In other words, instead of saying a solution doesn't exist, why can't we say two distinct solutions exist?
Because that's not the definition of a limit. Loosely speaking, if a limit exists, all of the functions values should "tend towards it" as x approaches 0.
That said, the one sided limits do exist.
Am i the only one that watch these types of math videos without understanding them? I'm in 8th grade and i have no idea of what this guy is talking about
9:14 Your x cubed term is wrong. You've taken x cubed term from sin but ignored x cubed term from log. I know x cubed term doesn't matter in the end but this has to lose you a mark.
Maybe, +∞ = -∞ = ∞
Huh?
Sir, when I saw trig functions like sinx,tanx and limit (xtends to 0).I considered them as x.This is very helpful in exam 😇😇😇😇😅😅
L'Hôpital Rule and not L'Hospital's Rule. It's a last name so don't change it!
I'm sure lindsay lohan could pull it off
asnwer=1 isit 🤣🤣
Rick & Morty
how the fuck is
ln(1-x)' = 1/(x-1)
Love the videos but absolutely despise how you draw your infinites