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?

(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.

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."

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 🙂

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?

…”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.

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

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.

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).

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.

if the value of cos(infinity) = 0, then the limit wouldnt exist because it would be 0/infinity. How do you know that cos(infinity) is not zero but another value between -1 and 1?

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

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?

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

OMG

L’ Hopital’s rule is NOT failing

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.

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

(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.

-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

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."

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

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.

ok

>Supposedly failure of rule

Still taught in school

This guy is quite annoying.

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

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

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

Ez

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

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

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

Sheldon

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

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

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

L'Hopital. Not Hospital

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

First of all, its L'Hôpital's rule.

So what ?

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. 😉

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

I corrected my pronounciation here

This is MIT?????

Oscillation of the function sin(x) around x-axis does mean, the function is trying to converge to 0.

So, it didn't fail.

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

i hate math its so pointless

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 🙂

I….don't agree.

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

I thought I am shitty at Math until I saw this

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.

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

Le hospital rule

Any JEE aspirants watching this video outta curiosity?

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

Did it 2 seconds…!!

Without using pen.🤣🤣

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?

Isn't this a little amateur for MIT UG ?

…”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.

Why does he look like he belongs on blues clues

I'm better than this guy.

Toooo bla bla bla

What do you mean by the second limit professor?

they wrote hospital in title

Limit as x tends to infinity of cos x doesnt exists

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

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

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

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

I did just this yesterday

Fail L'Hospital's Rule

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RATIONAL FUNCTIONS AND UNDEFINED FUNTIONS.https://youtu.be/36oC-ihvRKY

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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

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.

yea a failure.. just like me

OMG that is a big ass piece of chalk.

It applies only for 0/0

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.

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

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

👍

The limit does not exist.

such a snooze fest literally takes forever to make his point

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

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

X+cosx/x

1+cosx/x

1+0==1

Lol autocorrect to Hospital

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

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

Chiste

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

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

just a useful sidenote:

1 <– (x – 1)/x <= (x + cos(x))/x <= (x + 1)/x –> 1

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.

Professor : prove it.

Me : It’s trivial😆

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

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

Great video though

Really good video, thank you

Adrian Monk in an alternate universe

this was taught in high school in india

Damn, that chalk board is clean

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).

what an incredible instructor.

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

Huh?

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

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

Thank you so much sir

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

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

if the value of cos(infinity) = 0, then the limit wouldnt exist because it would be 0/infinity. How do you know that cos(infinity) is not zero but another value between -1 and 1?