Failure of L'Hospital's Rule | MIT 18.01SC Single Variable Calculus, Fall 2010

[@MrElectrototo]

"Constant but wigglier" - new favorite phrase

[@sdafsf]

Cristobal Liendo new definition of bounded functions

[@SODA63]

Cristobal Liendo funny af

[@PrashantBatule]

Tchebyshev approves 🌚

[@shermanthompson871]

So it didn’t fail. People just need to actually know the damn theorem.

[@seess8251]

Applying pythagoras theorem to non-right triangles and get wrong results doesn't mean the theorem is defective. A fridge is pretty useless if you want it to wash your clothes. I can go on... The error is in the user, who selected the inappropriate tool

[@chanpreet3000]

Absolutely, LH is used for only indefinite form .. the above mentioned was not indefinite

[@nickbroderick7595]

@@chanpreet3000 The original limit *does* have an indeterminate form when the limits of the divisor and dividend are considered respectively (+inf/+inf), which is one of the necessary conditions of the theorem. The other - that the quotient of derivatives has a limit - fails. That is the salient (and oft overlooked) point being made here.

[@prathamyadav3105]

Actually the proof is "proved" using tailor theorem. Proofs don't fail. They just require some conditions to be fulfilled.

[@98danielray]

@@prathamyadav3105 it comes directly from the definition of derivative

[@davidjames1684]

Maybe if you had the proper health insurance, L'Hospitals rule wouldn't fail.

[@marketmail49]

🤣🤣🤣

[@mikeljanhajrullaj5784]

Fail, fail, when you have to do with the square root! ...

[@ethanjensen661]

Lol the autocorrect

[@ryanadams9584]

Wooooooow

[@megauser8512]

lol

[@ivan1793]

No wonder the guy's smiling so much. I would too if I had such good quality chalks and board!

[@aashsyed1277]

nahhh hagoromo chalk is better

[@tdpro3607]

@@aashsyed1277 not the best, just good enough

[@pandit-jee-bihar]

If you end up using L'Hopital for limits then it's a good idea to keep on checking the indeterminate form at each stage and it must be either 0/0 or ~/~ to apply it further. In other word limit must exist for the rule to apply.

[@RaviSingh-qd5pz]

The limit does exit at stage second as it is not 0/0 or ~/~. You're contradicting yourself.

[@2013Aleksey2013]

"... or 0*∞." Otherwise your comment sums it up perfectly.

[@darkseid856]

@@RaviSingh-qd5pz he's not contradicting himself . He's just saying the conditions under which we can use L'Hospital Rule . And he is correct . It Requires 0/0 or ∞/∞ form to use this rule

[@yashagnihotri6901]

@@RaviSingh-qd5pz hehe , beginners

[@agustinsaenzanile1900]

​@@darkseid856why you can use it when you have inf/inf? i thought you only can use it when is 0/0

[@georgesadler7830]

From this short lecture, I finally understand L'Hospital's Rule from start to finish.

[@jaggysmf1860]

Damn, that chalk board is clean

[@atulchahalhr1657]

That the nice one😂

[@Vonzi0000]

@caoutchoucmilisec -1

[@chessandmathguy]

yep, well said. this is correct.

[@alphamega3306]

Nice to see the inclusion of the formal method showing that the limit as x -> infinity for the function cos(x)/x is equal to zero.

[@tino_S]

Are we assuming that x will be going to positive infinity or negative infinity here?

[@darkseid856]

@@tino_S doesn't matter the value of cos(x) ALWAYS lies between -1 and 1 (including -1 and 1) for all values of x

[@ocococoe]

The squeeze theorem shows that cos(x)/x = 0, but this question isn't asking for cos(x)/x. It's (x + (cosx))/x = x/x + cos(x)/x = 1 + 0 = 1.

[@mr.lonely1213]

R u alive ???

[@ProfessorDrGoku]

Paras Mittal nah he dead

[@darkseid856]

@@mr.lonely1213 it's been 6 years .... Obviously he's dead.

[@cipher01]

Been there since 2009

[@zwelakhe22]

He's right though

[@Giganesh_exe]

I would have used the squeeze theorem for cos(x)/x to show why it’s limit equals 0. Great video though

[@toastersman217]

You could have also squeeze the presented limit between (x-1)/x and (x+1)/x (By using |cos(x)|

[@anmoljhamb8775]

@@toastersman217 oh yeah, nice

[@bridge5189]

Conclusion: L'hospital rule is valid only when the limit of the first derivative exists

[@diegoarana5862]

Not really. If the first derivative doesn't have a limit you might be able to take the second derivative which might have a limit. It only didn't work in this case because the function had that pesky cosx.

[@bridge5189]

@@diegoarana5862 So, you mean one can go on and on having derivatives until you reach a point where the limit exists?

[@darkseid856]

@@bridge5189 yes .If the first derivative doesn't have a limit , then take second and so on.... But make sure to check the limit form in each and every step . Because L'Hospital Rule can only be used when we have a limit of 0/0 or ∞/∞ form .

[@Shaymin1010]

@@diegoarana5862 No! If the ratio of the first derivatives is a limit that doesn't exist, then you can never ever say nothing about the original limit - even if the ratio of the second derivative does exist.

[@Datboy1991]

diego arana Be careful with the terminology, because how this is phrased will impact the validity of the statement. A second use of L’Hosp is only warranted if you get an indeterminate form again, which is very different from saying the limit does not exist. If the limit actually does not exist, L’Hosp is invalid and you need to do something else.

[@MrQwerty15ification]

I really enjoyed this video. I never knew of that stipulation than the limit had to exist. When I worked out the problem I immediately tried to see if it diverges or converges. But it did neither. Very interesting!

[@askjfjawjhwahwjehjsjah1668]

L'Hospital's Rule never failed, it's your way of using it that fail.

[@ralphdams7568]

Ow man, this video was posted almost 9 years ago before it showed up at my side. :)

[@AlexssandroMeneses]

For me too!

[@primephoenix1.077]

Now its 10 years....😅

[@BizVlogs]

Could also use squeeze theorem. The function is greater than or equal to (x-1)/x ,and less than or equal to (x+1)/x

[@kevingast4519]

Love it. A professor at MIT is using chalk and a blackboard to explain math while public schools insist that every teacher use a $5000 Smartboard because kids can't learn with chalk. Love it. kg

[@wulphstein]

Damn! That was the one rule whose name I could remember!

[@Metalhammer1993]

it got butchered to "the hospital rule" all through out our course back then. Next yeat our professor had it, he introduced it like this. "Next we´re going to discuss L´Hopital´s rule. or the hospital rule as this is where somebody will end up if i have to hear that for another year"

[@Mordenperson]

Everyone saying this isn't correct is wrong as the question is the lim of 1+(cos(x)/1) and using squeeze theorem you get cos(x)/1 is 0 so the limit is infact 1.

[@hbarudi]

Limits can be separated through addition, subtraction, multiplication, and division. That rule shows up in the textbook before L'Hopital's rule.

[@tetraedri_1834]

Provided that the separated limits exist and are finite, just as in the l'Hospital. Two counterexamples, if we take limit to infinity, we get lim (x-x) =? (lim x) - (lim x) = infty - infty, lim(sin^2(x) + cos^2(x)) =? (lim sin2(x)) + (lim cos^2(x)) DNE. The first limit is obviously 0 and second is 1, but separating sums gave nonsense.

[@hbarudi]

@@tetraedri_1834 I agree, best to solve for functions that have a limit as x approaches what we are looking for before taking the limit itself.

[@tdpro3607]

lhospital is used if you compute a lim thats 0/0 or inf/inf, which is unable to obtain thru normal means

[@bek_quereshi]

So it is not L - Hospital's rule, It is lohpital's rule.

[@saburousaitoh]

Please look the paper: [29] viXra:2001.0091 submitted on 2020-01-06 17:52:07, (58 unique-IP downloads) Division by Zero Calculus for Differentiable Functions L'Hôpital's Theorem Versions

[@AnuragVaibhavv]

If we replace x by 1/t and change the limit to t -- 0 then we can solve that

[@anmpir]

what an incredible instructor.

[@Vonzi0000]

you can use the squeeze theorem to prove that the limit of cos(x)/x = 0

[@Red-Brick-Dream]

No, you can't.

[@christianondo9637]

I don’t see why you need to

[@danielgates7559]

MoonlapseVertigo he actually can, but it’s overkill...

[@isaacaguilar5642]

Jerry Tan yea theres no need in this case

[@andresmoraga1888]

MoonlapseVertigo Agreed. Nice name btw, my favourite song

[@ShreyaPalak]

We need more of these!

[@wesolyfoton]

Oh... but I can use L'Hospital rule to solve this limit! (x - 1) / x

[@michalbotor]

funnily enough l'hospital's rule doesn't apply for the ever more interesting case of the limit of this expression as x goes to zero either, as the numerator does not go to zero as the x goes to zero. quite an interesting function it is.

[@ny6u]

the cosine taylor series term divided by x goes to zero while x/x goes to 1 as x-> infinity

[@juanpedro19840914]

There's an extra "s". It's just L'Hôpital.

[@tsujimasen]

Juan Pedro Romero Either way is fine, yours is the older way, the s is the newer way.

[@HighLordSythen]

In old French, it was written with the s. Later, the ô replaced the os in French words like hospital >> hôpital and coste >> côte. This applied to other vowels as well. That's why you see it written both ways in modern materials.

[@pandit-jee-bihar]

Everyone in higher secondary makes that mistake the first time guess in my experience :)

[@SimsHacks]

L'hospital or l'hôpital.. both are possible. The accent remplaces the S.

[@yaoooy]

The funny thing is marquis de l hopital just stole that rule from another mathematician

[@michalbotor]

just a useful sidenote: 1

[@Dhruvbala]

Alternatively, one could use squeeze theorem as -1

[@tdpro3607]

its just a demonstration, a simple but bad one since it missed the crucial point of using lhospital, for when a limit seems to reach 0/0 or inf/inf.

[@tdpro3607]

3b1b's explanation should help idk

[@pathagas]

so L’Hospital’s rule can’t prove that a limit doesn’t exist? this completely changes my entire Calc I experience.

[@jarikosonen4079]

The 1-sin(x), x->Inf doesn't make sense. It is possibly good to note that LH wouldn't work backwards. So to say that if '1-sin(x)' was given task you'd integrate it backwards (cause if its not proven its not LH pre-results) and then do necessary steps and get result of '1'. LH doesn't work for sine and cosine infinite cases and nothing works, except the DNE (does not exist).

[@ffggddss]

Let's see . . I can think of a couple ways L'Hôpital could be said to "fail" - 1. You violate its premise (numerator & denominator must both → 0 or both → ∞) 2. The premise is satisfied, but the rule keeps yielding a result that needs for the rule to be applied again, thus never ending on a final result Will it be one of these? I see from the opening expression that there's a 3rd way - 3. The premise is satisfied, but the rule yields a divergent (non-converging) expression when the original expression converges So now the question becomes, how do we guard against this circumstance, in a general case? Post-vid: OK, so there's a 2nd premise - the resulting limit must exist. Important to note also (to emphasize a point made in the video), that sometimes, when the result DNE, the original also DNE. Fred

[@nimeshpoudel89]

1st question limit is 1

[@georgechris420]

That is an logical explanation but what also exists is the "sandwich theorem"as we used to call it in class, which says if limf(x)

[@IshaaqNewton]

Oscillation of the function sin(x) around x-axis does mean, the function is trying to converge to 0. So, it didn't fail.

[@vikramvalame9990]

No

[@IshaaqNewton]

@@vikramvalame9990 No man, I used Probability Distribution to find it. It's not a fairy tale.

[@Jeremy_Fisher]

@@IshaaqNewton Sure, but using L'Hospital's Rule here doesn't get you any closer to the answer than not using it, so in that sense it failed in it's duty to help you find the answer.

[@IshaaqNewton]

@@Jeremy_Fisher Think again. lim(x->0) sin(1/x)=0 it also helped me to find integral of cosx/(x^2+1) between 0 and infinity while using Feynman's technique.

[@2013Aleksey2013]

Here's another approach: lim(x --> ∞) [(x + cos(x))/x] = lim(u --> 0) [u/u + u*cos(1/u)] = 1 + 0 = 1. 1/x = u; x = 1/u. -1 ≤ cos(1/u) ≤ 1.

[@aartivishwakarma7081]

But the lim ₓ→ₙ 1+cosx/x where n=∞ , will be 1+∞/∞= undefined not 1.

[@volodask]

cos x does not have a limit in infinity. As Joel says, it behaves like a "wiggly constant", so the limit is approximated as 1 + c/∞, which is 1.

[@justviewer5458]

Use squeeze theorem to solve Lim x to the infinity , cos (x)/ x

[@joaopalrinhas5242]

You can only use the L'Hopital rule when there is an undertermination, you just can't define a limit to cos or sin functions to infinity.

[@abrahammekonnen]

Great video explaining the topic in further depth. I would recommend doing this more often if you haven't already

[@mancinellismathlab7451]

Great explanation and example

[@BobbyJCFHvLichtenstein]

I did just this yesterday Fail L'Hospital's Rule

[@merveilmeok2416]

...”So why don’t you pause the video, spend a couple of minutes to work on that and and we’ll are going to work on it together”. I heeded the advice and I got a hamburger from the kitchen.

[@tapaskumarpaul4982]

find the limit (f^-1(8x)-f^-1(x))/x^(1/3) as x is tending to infinity where f(x)=8x^3+3x Please solve it

[@supertren]

We must to check this three statements: 1) f(x)→±∞ 2) g(x)→±∞ 3) f′(x)/g′(x)→L But the 3 limit does not exist, therefore we cannot apply L'Hopital's rule.

[@mrAmal45]

In my college it is L- Hospital's rule not Lopital's rule

[@henri1_96]

I think it's just pronounced "Lopital"

[@alimehrabifard1830]

Change x to 1 over x and then the x goes to zero (equal to the initial condition where x went to infinity), you get 1 super quick. ;)

[@yaoooy]

Yes but change nane to the variable. X cannot be substituted by 1/x but rather by 1/sonething else, just a formality :)

[@metarus208]

The limit does not exist.

[@marcosmimenza]

So that's how a behind the scenes obsolet green screen looks like without the movie on it?

[@Hyajae]

The chalk reminds me of the times where I drew on the sidewalk with chalk. So yesterday.

[@Eduardo-hv1yh]

Adrian Monk in an alternate universe

[@JeroAlmufakir]

OMG

[@energy1136]

Don't you actually need to proove that lim(cosx/x) as x spproaches infinity is 0 with the use of the sqeeze theorem?

[@inakibolivar664]

TheFiendMegaCyber97 No, we just need to apply the constant but wigglier rule and done

[@Jungleland33]

Why don't I not pause the video.......... instead I'll watch this with a puzzled look on my face and my jaw on the ground.

[@morschlesinger7081]

It's Jerry from Rick and Morty

[@crocopie]

Omg! Even the shirt is the same! I'm glad Jerry in our universe teaches at MIT.

[@Usuario459]

@@crocopie this Jerry would be a moron?

[@CodyShell]

why can't my recisitation teacher be this good?

[@mr.lonely1213]

R u still on utube?

[@ralphdams7568]

L’ Hopital’s rule is NOT failing

[@keineangabe8993]

Right, it's just not applicable. Big difference.

[@yaoooy]

Nope it is applicable if you change variable

[@chrismanson3211]

I teach calculus and I did not know that fact, thank you.

[@dhruvchaurasiya545]

Limit as x tends to infinity of cos x doesnt exists

[@Jeremy_Fisher]

Good thing the problem isn't limit as x tends to infinity of cos x

[@adityatiwari2488]

Who came here after the post of Jee simplified

[@Kartik-yw7kp]

me 😃😃

[@ガアラ-h3h]

Welll obv the limit is just one cause you can split the fraction and then cos x: R -> [-1;1] this you get 1

[@bobkameron]

Great video and explanation!

[@SherlockHolmes-di2yr]

2:40 it's never near negative 1 pal

[@Vibranium375]

Uhh isn't sin X always between 1 and -1? For example sin (3π/2) = -1

[@SherlockHolmes-di2yr]

@@Vibranium375 I meant 1-sinx is never negative 1

[@Vibranium375]

@@SherlockHolmes-di2yr oh ok I misunderstood ur statement :)

[@pascaldelcombel7564]

You should review what is really L'Hospital rule before saying it fails...

[@sujeetkumarroy3062]

Ya its good as you tought ...😍😍😍😍

[@swatisingh1316]

Thank you so much sir

[@melchiortod29]

Really good video, thank you

[@lleo3638]

The rule actually says, if lim(f'/g') converges and equals L, then lim(f/g) also converges and equals L, hence the equation lim(f/g) = lim(f'/g'). In this case, the rule doesn't fail, the conditions are not met at the first place.

[@devanshsharma743]

we'd not use l'hopital rule in this, because it's not in the form of any number divided by zero

[@tsujimasen]

Devansh Sharma you can use l'hospital for 0/0 as well as inf/inf

[@plaustrarius]

Leh hos pi tal

[@davidjohnston4240]

Where did that nomenclature "the second limit" for "the limit of the derivatives" come from? I've never seen that before.

[@agustinl2302]

I'm pretty sure it's not standard nomenclature and he was just informally referring to it as the second limit that would come up. Just like he explains cos(x) as "a constant but wigglier".

[@7ymke]

even mit says its L Hospitals instead L'Hopitals

[@positivegradient]

The limit of periodic functions like sin and cos does not exist as we go to infinity, since it's an oscillating function, and infinity is not some exact number, so we don't know what value sin or cos will take.

[@LaneSurface]

But that's not what we are taking the limit of. It's the lim(cos(x)/x), which does approach a value (zero) as x tends toward infinity. x>>cos(x) as x->inf.

[@positivegradient]

@@LaneSurface You're right, but the expression you get by applying L hospital's is (1-sinx), the limit of which does not exist. Even though the original expression tends to 1 as x goes to infinity. This is why L hospital's rule fails here.

[@positivegradient]

I was trying to highlight that you need to be wary of periodic functions in x goes to infinity limit problems.

[@SriNiVi]

I wouldn't call it failure per se.. you just can't use it in this case and it's illogical to use it too.

[@zwelakhe22]

I thought I am shitty at Math until I saw this

[@pandit-jee-bihar]

Not all limit problems are recommended to be solved using L'Hopital rule.

[@samuelprakash5734]

You guys do know that it is L’Hôpital’s rule not L’Hospital.

[@nelsonmcnamara]

Lo-pi-tal's rule. LOL. Been saying it as "La Hospital" Rule all my life....

[@rockspoon6528]

Why would you bother using L'Hospital's rule on an equation that so obviously trends to 1...

[@santhoshwagle9857]

What do you mean by 'second limit exists'? How do you say second limit doesnt exist in this case??

[@heramb575]

What do you mean by the second limit professor?

[@vikramvalame9990]

The limit one gets by taking the ratio of the derivatives

[@heramb575]

@@vikramvalame9990 ok ..why does it need to exsist though?what's the proof for that?

[@anushkaartsandcrafts3069]

L hospital rule used in limits to solve the problem easily this is not part of limit so it just like a trick so I think this rules never fail

[@UjwalAroor]

gyanenendr pandey It is not a trick.L’hopital’s rule is based on the assumption that once you differentiate it,the new limit must be in inderminate form.Most beginner classes dont teach this but that is an assumption made by the rule.Most of the limits where you applied l’hopital’s rule were in indeterminate form.The above problem is not a trick,it is just an exception.

[@stiventson4464]

@@UjwalAroor I don't get it, you wanted to say that the limit must be indeterminate when you just *evaluate it* to it to work? because doesn't make sense that when you differentiate the limit, it must be inderterminate... I mean , On the contrary, it should be determined so it would works...

[@UjwalAroor]

Stiventson44 Ok so i'm not exactly sure what you are trying to convey with your comment.But as I have interpreted it, I think you are trying to say that when we differentiate a limit, it must always be in indeterminate form.This is not and the video above is proof.But if you are not saying that then please reply to this comment.

[@stiventson4464]

​@@UjwalAroor this line you wrote "L’hopital’s rule is based on the assumption that once you differentiate it,the new limit must be in inderminate form" I don't know why you said the new limit must be indeterminate once you differentiate

[@UjwalAroor]

Stiventson44 Dude if the new limit is not in inderminate form then most of the time you cant solve it.In the above example,the limit wasnt in indeterminate form so it couldn't be solved.

[@DjVortex-w]

That was the least mathematically rigorous calculation I have seen in a while.

[@HarshRajAlwaysfree]

Maybe cuz I'm just in High school my mind was blown!!! I will remember this thing till death who knows when someone gives this question to trick me :)

[@CounterTheAnimatorocn1]

QUICK, What is the limit of (x+cosx)/x as x approaches infinity?!!??!

[@z1lla4]

@@CounterTheAnimatorocn1 didn't answer, guess he wasn't ready

[@CounterTheAnimatorocn1]

@@z1lla4 maybe he died

[@SuperParaNatural]

@@CounterTheAnimatorocn1 oof.

[@gauravmishra1508]

Very easy question. Answer is 1.solved by geometry while pausing your video while you told so at 0:51

[@saisitharth]

It applies only for 0/0

[@lunkel8108]

No, also for infinity/infinity. Both are equivalent, really. Just take both the numerator and denominator to the power of -1 and it's plain as day.

[@prasadsawant1358]

they wrote hospital in title

[@dendion711]

cosx/x goes to 0 as x goes to infinity because: |cosx|

[@yaoooy]

You can't apply hoptal rule on trig functions that tend to infinity

[@michaelbayer5887]

... its Opera or Operation like calculation - as OPERA and COSIMA fan tutte.

[@vibration1028]

I corrected my pronounciation here

[@ultimatevideoproducer2117]

Very good explanation. Also your handwriting is neat. One thing I'd improve is how you explained "dividing by x" to the x+cosx/x to get 1+cosx/x... I only followed your explanation of that since I've somewhat done this before.

[@protodosto]

x/x = 1

[@notsoslimkids]

Why couldn't this be solved with the squeeze theorem? With that method, the limit as x approaches infinity is 0. So, is the Squeeze Theorem incorrect in instances?

[@Mordenperson]

The question is Lim of 1 + (cos(x)/1)

[@071-babum4]

Can you change the title please..?? L'Hopital rule is correct one.!!!

[@rahulbag2931]

Why does this limit not exist for sinx but exist for (cosx)/x though both are oscillating function?

[@exploremaths6138]

nice sir .

[@rc_youtubeaccount1331]

a shaking constant....fck why i didn't see that

[@canalf007]

-1

[@subhasdh2446]

I like burger more 🍔

[@amitir22]

why? it's just lim1+ lim cosx/x lim 1 = 1 lim -1/x

[@canalf007]

@@amitir22 you cant just separate the expresions like that.

[@danielnewby2255]

@@canalf007 Yes you can. It's just an application of the sum rule.

[@canalf007]

@@danielnewby2255 separate lim (1-cos x) / x^2, as x tends to 0

[@Emerald_Solace]

Why does he look like he belongs on blues clues

[@kaustavchakraborty9484]

Isn't this a little amateur for MIT UG ?

[@Topspeedcraft]

XD bet this was in the same class the diff calc syllabus was presented