Hopital rule proof

alright thanks for watching and today I will prove the celebrated l'hopital's rule you will see the proof is really beautiful it brings us through some analysis but it's one of the proofs so you have to be extra careful however your constants are and there will be lots of pain lots of Epsilon's but in the end you left with a very neat epsilon at the end so and it also leads us to a generalization of the mean value theorem which is also super super cool so I really can't wait to start it but first of all this one thing about l'hopital's rule dopey todd didn't actually come up with it it was supposed to be Bernoulli's rule but dopey Tao at that time had stuff mathematicians already have lots of money so he literally bought you know l'hopital's rule so i guess money can do everything ok anyway we are more happiness that's all that matters so remember what does lupita rule say so you know a lot of popular students they just write blindly nor limit X goes to a f of X over G of x equals to limit X goes to a of f prime of x over G prime of x equals to L to find out how it's easy the limit code equals to L but what does that really mean it means that if this limit exists and equals to L then in fact the original limit exists and equals to L so let me write that down in more detail so 0 so suppose F and G are differentiable are differentiable and let's just use the blue rule like zero over zero because the rule infinity over infinity it's just an adaptation of this proof so if F and G are differentiable and the limit as X goes to a of f of X equals to zero limit X goes to a of G of x equals to zero in other words against suppose you of the case there are over zero and suppose that the ratio of the derivatives X goes to a of f prime of x over G prime of x equals to L then we have that the original limit f of X over G of x equals to L again this is funny because when calculus student doesn't prove like that you know red solves a problem they actually go through that whole process unconsciously but they don't really think about that but again this is the statement of l'hopital's rule and in this proof again otherwise it becomes more complicated let's just assume that G & G prime you're not 0 it never 0 or something but and as I said it's a really nice analysis proof so if you like analysis this is for you so step one we want to show this right that the limit of f of X divided by G of X is M so using epsilon and Delta is what does that mean so oh sure limit X goes to a of f of X over G of x equals to L so let epsilon be a positive and we want to find its want to find delta positive such that if X minus a is less than Delta then the ratio f of X over G of X minus L is less than Epsilon okay well so remember when we show f of X over G of X minus L if at the end with 1 epsilon we're happy this remember epsilon means happiness all right now let's fix X okay so for now think of X as a constant like 2 6x and net he is to be announced he will depend on X and there will be to be announced and what's the idea notice if you think about this F prime of X over G prime of X well s prime think of it as a difference quotient so think of it as you know f of X minus F of T over X minus T and G of F prime of X is G of X minus G of T over X minus T and think of F of T and G of T are small so after this cancellation we get f of X minus f of T over G of X minus G of T I'm making a video oh oh sorry did not do well so we have this conference oh okay no we don't need it we don't need this room but if anybody comes here okay okay sounds good yeah yeah all right there where we go that f of X minus f of T divided by G of X minus G of T which sort of modem just want to motivate why we need that ratio and what follows so remember what we're focused on f of X over G of X minus L we want to show this is small wanna show this is less than Epsilon and now this is a nice math trick that appears over and over again analysis and I just want to say if you're ever stuck in analysis this is probably why because you forgot that trick namely we want to take this quantity and I don't subtract that quantity so minus f of X minus f of T over G of X minus G of T and we want to add that back to make it legit f of T over G of X minus G of T minus L so think of it as a reverse cancellation if you cancel that out then you do get that as equals to f of X over G of X minus L okay good so it's kind of weird because we took something easy and made it horribly complicated and in fact one of the things about analysis we want to show this is less than epsilon so let's estimated so by the triangle inequality this is less than equal to f of X over G of X minus f of X minus f of T over G of X minus G of T absolute value parentheses plus f of X minus f of T over G of X minus G of t minus f so we're left with two terms and for the next half an hour or so we will study those two terms let's call it a and let's call it B I notice we still don't know what he is so so far everything works with general T and again remember our goal is to get an epsilon at the end so we wanna show that those terms are small which again it makes sense remember later and this will be small and this will be small so this whole difference should be small and this is sort of like the derivatives and we know that the derivatives they go to L so we're good alright so step three study of a remembered a is that term here so f of X over G of X minus f of X minus f of T over G of X minus G of T okay so this is a difference between two of two ratios there's nothing else to do than just to put it on a common denominator so this is f of X times G of X minus G of t minus f of X minus f of T and G of X is absolute value over the products of G of X and G of X minus G of T I'm not so excited just know when you see something beautiful you just like you want to share it okay let's expand it out so we get absolute value f of X G of X minus f of X G of t minus G of X f of X plus G of X f of T divided by G of X G of X minus G of T now one nice thing is and this is what we expanded it out bang-bang f of X G of X and G of X f of X they cancel out and you're left with this term minus f of X G of T plus G of X f of T over G of X G of X minus G of T hipster graph because G of T it rhymes me of DT and it is my cousin's name et Cie to be one-sided okay French Switzerland so now the numerator again there's nothing really we can do but remember we want to estimate stuff so let's just use a very rough triangle inequality so this is less than absolute value of this plus the absolute value of this so it's f of X G of T plus G of X f of T / G of X G of X minus G of T and this is the partner was a bit stuck on and to be honest I did look at the solution because it's not quite obvious but if you think of this this is sort of two parts of a foil expansion so if you take some polynomial and you expand it out you get those two terms but also two other ones that are missing and in fact it turns out that polynomial is f of X for that term and for that space G of X f of T plus G of T divided by G of X over G of X minus G of T so notice if you expand the top out those are precisely the cross terms f of X G of T G of X F of T the other turns red and form X f of X is f of T G of X G of T but they're all positive so you just taking this expression adding more positive terms okay why do you do that again later as I said show that F of T and G RT are small so those two things will be very small this will be constant this will be constant and we'll see this will always so be a constant so it's sort of this is the controlling part of the term a and remember we wanna show a is small so we are good hands and by the way this proof is taken from real analysis book which is one of my favorite books so I highly invite you to read it okay now this thing it works for India well any X in this region but also for any tea and now I can finally release you of your torture and tell you what he is so given X again X is a constant think of it as like two or something let t be something the following let t be so much such that maybe let me draw you a little picture okay we have this X we have this a right X is very small to a a want t to be even closer to a let you imagine for example this is fire and like I'm pretty close to the fire but I'm kind of scared to go but hey who says fire says have flammable mass so flammable mass is like I'm not scared I will protect you so this is sort of what's called an advance guard or something so it's a number that's even closer to a and such that the following truth is true namely F of T plus G of T it remember it was a small F and G they go to 0 as T goes to a and particularly picton such that this is smaller than okay don't laugh G of x squared over 4 f of X plus G of x times Epsilon ok again how in the world did I pick it it's you'll see at the end it makes the things cancel out so for example this will cancel this one what factor of G will cancel this one and we need some other terms but again I want to emphasize remember that X is fixed so what this says it just says assume that F plus G is smaller than a constant times Epsilon okay so think of three Epsilon and that's fine because remember that at F of T and G of T go to 0 F of T goes to 0 and G of T goes to 0 as T goes to 8 that's one of our assumptions and then we'll also choose another condition to be announced so choose T such as this condition holds and an extra condition will hold we'll try write down in maybe five minutes or so okay once we have that let's play around with this so as I said a is less than equal to f of X plus G of X over G of X and then G of X minus G of T now this term hallelujah it actually appears exactly in this expression so this expression is less to equal to G of x squared over 4 f of X plus G of X and now again here's our math is really cool lots of terms cancel out so this term cancels out with this term one of the factors of G also cancels out and again if you didn't know what that constant was just say pick you know T such that this is less constant times epsilon and the idea of a concept has Epsilon but it's super interesting because notice also the interplay between X and T so we really have to judiciously choose our concept so so in particular a is Leicester you go to epsilon over for G of X over G of X minus G of T okay and now that's not that ideally we'd like some constant so we have to work a little bit harder maybe to make this G of T disappear and ideally also that G of X disappear and for this we need a second condition on T namely remember this is small in particular we can choose it to be less than G of X over 2 again this is a constant so in other words make G of T small and the nice thing is we have some my estimate here's the cool thing usually you want absolute values you want to use the triangle in evolved quality you want to show it's less than something but because you have a reciprocal you want to show it's greater than something so you have to use what's called the reverse triangle inequality which remember it's just as an apse value of a B minus B is greater equal to absolute value of a minus absolute value of people in particular this is greater or equal to G of X minus G of T and now G of T is less than that so minus G of T is greater than that so G of X minus G of X over 2 but that's just G of X over 2 in particular let's go back to a a is less than equal to epsilon over 4 G of X now you're taking the reciprocal of this because you have 1 over that and so you take 1 over G of X over 2 so 2 over G of X look at that this cancels out and you left with epsilon over 2 absolute value a is Leicester if I'm strictly less than and so all that up that said I've some bad news we've just done half of the work because now we want to analyze B but I think B isn't as bad that's all well see okay a is less than epsilon over 2 if only B were less than epsilon or two we can manage that so it's step four study of B and you'll see this this is a video that actually makes you smarter so we're not only doing analysis proving new stuff too because what is B member B may be an absolute value but the difference quotient right f of X minus f of T G of X minus G of T minus L okay and here's the idea remember what the mean value theorem says it says that the top can be written as f prime times X minus T so by the mean value theorem this term over this term minus G of T well the top you can write it f prime of C times X minus T over G prime let's say of T times X minus T this cancels out and you're left with F prime of C over G prime of D and I want to emphasize it doesn't necessarily give you the same number right we know that for some C this is true and for some other D resistor what if C is equal to D so if C equal to D then we have then we get F prime of C over G prime of C but remember if Lexi goes to a again very hand wavy if C goes to a this actually goes to L because remember we've shown that we assume that F prime over G prime it goes to L so if on one hand C equals to D and you can let C go to a then this ratio goes to L in particular we can make it less than epsilon over two which requires again tooling outrageous things first of all 12 the same number in the mean value theorem and also to let this number go to a it turns out we can do this I know right like wishful thinking that is true so first of all can we guarantee that C equals to D yes and this is what's called the ratio B value theorem namely there exists a C in X T or T X whichever is in my picture T was less than X ovt X such that f of X minus f of T over G of X minus G of T equals to F prime of C over G prime of C in other words the ratio the differences can be estimated or is equal to the ratio of the derivatives pretty neat and why is it called ratio me value theorem if G is the identity function then this becomes X minus T G prime would be 1 and you get f of X minus f of T over X minus T equals to F prime of C so in fact the mean value theorem and let me quickly prove it it's very very deep it's just actually an adaptation of rules defy the following function feel us to be again everything is fixed so except for s so f of X minus f of T G of s s minus G of X I know people come up with those functions but minus f of X minus so minus G of X minus G of t minus G of t f of s minus f of X okay let's see what's happening so what is field and X well this becomes G of X minus G of X which becomes 0 this becomes f of X minus f of X which becomes 0 so the whole term is 0 and what about a T this is also super nice this becomes G of T minus G of X this becomes f of t minus f of X and so I believe that the terms and cancel out because you get for example you know - you know that anyway if you do that the terms do cancel out so V of T equals to 0 and so by rule we get that V prime of C equals to 0 and what does that mean again for some C X T but what is feet prime of s remember everything is a constant except this G so that is except s sorry f of X X minus f of T G prime of s and then minus G of X minus G of T F prime of s so 3 prime of C is equal to 0 what that means is that f of X minus f of T G prime of C equals to G of X minus G of T F prime of C so this whole term would be 0 so this term equals 2 adapter and then for example divided by G Prime and divided by G of X minus G of T and you get that the ratio has to satisfy the mean value theorem issue profit wonderful this was like the last hard step and now we're almost done I think you see a or like sad don't you want to see more adventures like that don't worry I have more videos so so what do we have well B was equal to f of X minus f of T over G of X minus G of t minus L now by the ratio mean value theorem that equals to F prime of C over G prime of C minus L and remember where C it was so remember that was a and then we had X and we are T and so see squeezed between X and T so and remember there was this Delta so this whole distance is Delta and because X minus a is less than Delta so it's kind of a funny story remember that was the flame that was me being scared that was like flammable masses like I'm not too worried and then that can read pen has like squeezed between the two like PI am I give you also protection nice semantics is so that's why guys like that I'm getting crazy it's like the song that preys upon that and so what do we have remember that the limit as X goes to a of f prime of x over G prime of x equals 2l and in fact if we haven't chosen our delta yet we said we want some delta such that this is true so in fact because we know this now we have our Delta so there exists a delta such that if X minus a is less than Delta then f prime of x over G prime of X minus L is less than epsilon over 2 in particular C satisfies that because we said key is even closer to a and C is between the two and prime of C over G prime of C minus L is less than epsilon over 2 and lastly just a quick conclusion with that Delta Delta what do we have f of X over G of X minus L its lesser equal to the first term a that involved in X and T and the second term with the ratios we've shown that this is less than epsilon over 2 plus x over 2 and here comes a satisfying moment this is Epsilon pretty neat and therefore l'hopital's rule is proven so this in that case I know a bit harder than you would have thought maybe but again it's just this beautiful excursion into analysis and this little story was like front guards rear guards so kind of cool and again all this goes in your head when you use l'hopital's rule so it sits on something very obvious all right so if you like this calculus analysis excursion and you want to see more mouths please make sure to subscribe to my channel thank you very much

Channel: Dr Peyam

Views: 31,549

Rating: 4.9061584 out of 5

Keywords: math, proof, calculus, analysis, famous proofs, hospital, hopital, hopital's rule, l'hopital, l'hospital, l'hospital's rule, l'hopital's rule, proof of l'hopital, limits, limit of quotient, epsilon-delta, advance guard, pugh

Id: tL13JxmyRLw

Length: 34min 41sec (2081 seconds)

Published: Mon Nov 12 2018