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

### 100 thoughts on “Failure of L’Hospital’s Rule | MIT 18.01SC Single Variable Calculus, Fall 2010”

• October 26, 2019 at 2:53 pm

Am i the only one who came here to see whats up with those cool letters ?

• October 26, 2019 at 8:54 pm

L’ Hopital’s rule really? Shouldn’t you be dividing the numerator and denomenator by x instead and then take the limit of x -> inf? This would reveal the limit of 1/1 = 1 right away, right?

• October 26, 2019 at 8:56 pm

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

• October 26, 2019 at 9:00 pm

L’ Hopital’s rule is NOT failing

• October 26, 2019 at 11:36 pm

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.

• October 27, 2019 at 2:56 am

That guy seemed to have a serious case of social anxiety.

• October 27, 2019 at 5:39 am

(1-sin(x))/1. As x gets infinitely large the sin(x) becomes insignificant and is negligible. Therefore you are left with 1/1 which gives you a limit of 1 which is the answer you were looking for. Therefore l'hopital's rule works in this example.

• October 27, 2019 at 5:41 am

-1 <= cos(x) <= 1
x-1 <= x+cos(x) <= x+1
(x-1)/x <= (x+cos(x))/x <= (x+1)/x
1 <= lim_{x–>inf} (x+cos(x))/x <= 1
then lim_{x–>inf} (x+cos(x))/x = 1

• October 27, 2019 at 7:56 am

By your method of explanation you are implying that that equality could make sense and that it is the L'Hospital's rules that are rejecting it. However, I think it really can't make sense. Do we really make an equivalence class of all failed limiting process attempts? I don't think so. Perhaps you don't need to touch upon the subtle point of the use/misuse of the equality, but you should make the explanation without reinforcing the false impression that such an equality makes sense.

In fact, if the L'Hospital's rules asserted the implication that if the limit of the ratio of the derivatives did not exist then the original limit also does not exist then I think that fact would have to be mentioned by itself in addition to the statement of equality. "…furthermore, if the resulting limit does not exist then the original limit does not exist."

• October 27, 2019 at 2:04 pm

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

• October 27, 2019 at 2:38 pm

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.

• October 27, 2019 at 7:35 pm

>Supposedly failure of rule
Still taught in school

• October 27, 2019 at 8:42 pm

This guy is quite annoying.

• October 28, 2019 at 12:33 am

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

• October 28, 2019 at 2:33 am

This dweeb is PRIDEFULLY teaching now at GWU. This guy is annoying. He was the worst out of MIT crowd.

• October 28, 2019 at 3:22 am

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

• October 28, 2019 at 3:44 am

Ez

• October 28, 2019 at 3:58 am

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

• October 28, 2019 at 5:01 am

When I broke my arm in Mexico, I went to the El Hospital.

• October 28, 2019 at 5:07 am

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

• October 28, 2019 at 9:31 am

Sheldon

• October 28, 2019 at 3:34 pm

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

• October 28, 2019 at 6:35 pm

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

• October 28, 2019 at 10:22 pm

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

• October 28, 2019 at 10:25 pm

L'Hopital. Not Hospital

• October 29, 2019 at 12:27 am

This is NOT an indefinite form since cos(0)=1 and not zero.

• October 29, 2019 at 12:57 am

First of all, its L'Hôpital's rule.

• October 29, 2019 at 6:10 am

So what ?

• October 29, 2019 at 7:05 am

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. 😉

• October 29, 2019 at 5:01 pm

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

• October 29, 2019 at 9:51 pm

I corrected my pronounciation here

• October 29, 2019 at 11:58 pm

This is MIT?????

• October 30, 2019 at 12:30 am

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

• October 30, 2019 at 3:47 am

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

• October 30, 2019 at 6:59 am

i hate math its so pointless

• October 30, 2019 at 11:04 pm

Or you could conclude that limx->ω(sin(x)) = 0. ω is being used to represent an ordinal infinity. Now ω is a large hyperreal integer and is thus divisible by all reals to get an integer result. Thus ω/2π is an integer so sin(ω)=sin(2π×Z+)=0. Thus limx->ω(cos(x)) = limx->ω(sqrt(1-sin^2(x))) = limx->ω(sqrt(1)) = 1. (Implying that e^ωi = 1 as well). Btw I realise this is completely unrigourous and I would like to know exactly why it is unrigourous if anyone is willing to explain 🙂

• October 31, 2019 at 4:23 am

I….don't agree.

• October 31, 2019 at 5:26 pm

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

• November 1, 2019 at 12:52 am

I thought I am shitty at Math until I saw this

• November 1, 2019 at 10:48 am

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.

• November 1, 2019 at 2:36 pm

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

• November 2, 2019 at 12:42 pm

Le hospital rule

• November 3, 2019 at 6:32 am

Any JEE aspirants watching this video outta curiosity?

• November 3, 2019 at 1:19 pm

Use u/v rule u/v=vu'-uv'/vpower2

• November 3, 2019 at 2:15 pm

Did it 2 seconds…!!
Without using pen.🤣🤣

• November 5, 2019 at 8:46 am

This is not for nitpicking or anything but just to be sure, at t=2.40, he mentions that "In particular (lim_{x to infty} frac{1 – sin{x}}{1}) it is near 1, sometimes near -1, sometimes 0". But, I think it can't be near -1, the range of values it can take is [0, 2]. Am i correct?

• November 5, 2019 at 11:52 am

Isn't this a little amateur for MIT UG ?

• November 7, 2019 at 6:52 am

…”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.

• November 9, 2019 at 3:47 am

Why does he look like he belongs on blues clues

• November 10, 2019 at 12:24 pm

I'm better than this guy.
Toooo bla bla bla

• November 10, 2019 at 1:12 pm

What do you mean by the second limit professor?

• November 13, 2019 at 11:42 am

they wrote hospital in title

• November 14, 2019 at 9:19 am

Limit as x tends to infinity of cos x doesnt exists

• November 14, 2019 at 10:36 am

Oh… but I can use L'Hospital rule to solve this limit!

(x – 1) / x <= (x + cos x) / x <= (x + 1) / x
L'Hospital:

lim (x-1)' / x' = lim 1 / 1 = 1
lim (x+1)' / x' = lim 1 / 1 = 1
Squeeze theorem:

lim (x + cos x) / x = 1

• November 18, 2019 at 1:20 am

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

• November 18, 2019 at 6:39 pm

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

• November 21, 2019 at 2:34 am

Nonsense title. It's not failing, that's just not the purpose of the rule

• November 21, 2019 at 11:20 am

I did just this yesterday

Fail L'Hospital's Rule

• November 21, 2019 at 5:06 pm

NEW VIDEOS ☺

RATIONAL FUNCTIONS AND UNDEFINED FUNTIONS .

https://youtu.be/36oC-ihvRKY

CONTINUOUS FUNCTIONS

https://youtu.be/QSFc4Q43OXE

MAXIMUM AND MINIMUM POINT, EXTREMAS.

https://youtu.be/-ggSkD3WcKE

• November 24, 2019 at 6:46 pm

To me sin infinity is 0 where we are taking the avg value of function in one complete cycle
Similarly sin^2 infinity is 1/2 due to its avg value in one complete cycle
And so on..
I don't know how can I proof it mathematically. But I know it's correct and one day I will proove it

• November 25, 2019 at 7:24 am

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

|cosx|<1 (by "<" I mean equal or less, there isn't a symbol for that on PC keyboard)
-1<cosx<1

-1/x<cosx/x<1/x -1/x and 1/x go to 0, so what is between them does so to.

• November 26, 2019 at 1:29 am

yea a failure.. just like me

• November 26, 2019 at 2:12 am

OMG that is a big ass piece of chalk.

• November 27, 2019 at 1:27 am

It applies only for 0/0

• November 27, 2019 at 8:01 am

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.

• November 28, 2019 at 9:35 am

engineering-and-science.com/calculus-for-dummies.html

• December 6, 2019 at 7:50 pm

I dont care if it fails. I want my stock holding to go up

• December 8, 2019 at 6:52 am

👍

• December 12, 2019 at 7:22 am

The limit does not exist.

• December 12, 2019 at 10:46 pm

such a snooze fest literally takes forever to make his point

• December 15, 2019 at 9:40 pm

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

• December 16, 2019 at 3:40 am

Alternatively, one could use squeeze theorem as -1<=cos(x)<=1 and then use l'hopital's to evaluate

• December 18, 2019 at 4:10 pm

X+cosx/x
1+cosx/x
1+0==1

• December 22, 2019 at 12:01 am

Lol autocorrect to Hospital

• December 22, 2019 at 12:06 am

75% of everybody in the comments calls it L'Hospital's rule. It's L'Hopitals's rule

• December 24, 2019 at 8:58 am

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

• December 26, 2019 at 5:40 am

Chiste

• December 27, 2019 at 6:10 am

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

• December 28, 2019 at 2:44 am

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

• December 28, 2019 at 11:27 am

just a useful sidenote:
1 <– (x – 1)/x <= (x + cos(x))/x <= (x + 1)/x –> 1

• December 28, 2019 at 11:48 am

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.

• December 31, 2019 at 8:57 am

Professor : prove it.
Me : It’s trivial😆

• January 1, 2020 at 4:39 pm

I solved it on my own and lost interest in the video but I will give you the view

• January 2, 2020 at 2:39 pm

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

Great video though

• January 3, 2020 at 1:53 am

Really good video, thank you

• January 3, 2020 at 6:59 am

Adrian Monk in an alternate universe

• January 3, 2020 at 7:49 am

this was taught in high school in india

• January 4, 2020 at 6:00 am

Damn, that chalk board is clean

• January 4, 2020 at 11:25 am

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).

• January 4, 2020 at 2:00 pm

what an incredible instructor.

• January 4, 2020 at 6:22 pm

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

• January 5, 2020 at 12:28 am

At some point I was afraid professor Lewis didn't come back 😲

• January 5, 2020 at 2:42 am

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

• January 5, 2020 at 3:58 pm

Thank you so much sir

• January 5, 2020 at 6:08 pm

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

• January 6, 2020 at 1:49 am

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