Nice L'Hôpital Problem

A near exercise using L'Hôpital's rule. If the limit of f(x) + f'(x) exists as x goes to infinity and the limit of e^x f(x) goes to infinity, then show that f'(x) goes to 0. This seems hard at first, but is quick to prove, once you see the trick. Enjoy!
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[@adithyan9263]

Dr Peyam at 2:17 : use the - (ploduct rule(nah that ain't good👦)) Dr Peyam proceeds to: "prada lou"🧔

[@DrWeselcouch]

That was awesome! No better way to start off the morning than a nice proof!

[@stevenwilson5556]

I love the smell of functional analysis proofs in the morning…

[@thedoublehelix5661]

At first I was like these are such random assumptions

[@qschroed]

we had this one as an exercise 2 weeks ago

[@nanangmuhammad2067]

Great solution Mr. Peyam. 😀😀

[@galo5818]

Say that some limit exists is the same as saying that it is finite or not?

[@drpeyam]

Yes

[@thohoduc5987]

Love your videos! BTW, I think the condition e^x f(x) -> inf is not necessary because the L'Hospital only requires the denominator (e^x) -> inf. I also think the main motivation of this trick is that this differential limit is of 1st-order so e^x is exactly the integrating factor of it. This trick does not apply for 2nd or higher order ones, for which we have to use other techniques.

[@erfanmohagheghian707]

You can't set the limit equal to the result of L'Hopital. You should first find the latter and if it existed, it would be equal to the original limit. I refer you to the theorem! Contradictory example: sin(sqrt(x)) /sqrt(x) Needed to add for the theorem to be applicable, the denominator going to infinity is sufficient

[@PeterBarnes2]

I wonder if it'd be possible to break this if you had some functions where you can't combine the limits at the end there. I'm not sure what utility that'd have (maybe some shenanigans with integral transforms, I guess), but I think that'd be pretty neat.

[@stefanocarini8117]

Please let "what time is it? It's l'hopital time!" be a 2021 meme! dr peyam just rocks :)

[@sayanmaji2845]

Just awesome math...I love it...

[@IISH4RKZzII]

more! more! more! more!

[@eliyasne9695]

*WOW* That's neat!

[@dbmalesani]

If I may add an observation, I don't think one needs the hypothesis that lim eˣ·f(x) = ±∞ for x → +∞ to prove the thesis: as far as it is known that lim[f(x)+f'(x)] exists and is finite for x → +∞, then lim[f(x)] = lim[f(x)+f'(x)] and lim[f'(x)] = 0 for x → +∞. One can just go through the same steps, by considering f(x) = [eˣ·f(x)]/eˣ and then applying the de l'Hôpital rule. In fact, while in the [0/0] form of the theorem both numerator and denominator must be infinitesimals, the only requirement for the ∞ form is that the *denominator* tends to ∞ - the numerator can be anything. The Wikipedia page on de l'Hôpital's rule briefly mentions this.

[@dbmalesani]

See also this pretty article: www.jstor.org/stable/2689813 for a generalization.

[@pbj4184]

Neat indeed!

[@GuyMichaely]

Is the proof that (lim x->inf)(f(x))=(lim x->inf)(f(x)+f'(x)) dependent on the assumption that (lim x->inf)(f(x)+f'(x)) exists?

[@drpeyam]

Yes definitely!

[@TrueBagPipeRock]

so good

[@francaisdeuxbaguetteiii7316]

Excellent video Doctor!

[@gaeb-hd4lf]

how i love lopita lu

[@stevenwilson5556]

This is where a larger board helps, since I forgot that we were computing the limit of f ' (x) when we got it equal to the limit of f ' (x) + f (x) allowing us to get that lim f ' (x) = 0. A larger board would allow the original limit to still be visible when we get it equal to f ' (x) + f (x). I had to rewatch to see what had happened.

[@medsaifeddinekamoun2921]

Hi Dr.Pyem ! I had to pause to think about the problem and a very strange idea came to my head that I couldn't dismiss even after watching the video, and that is : if the limit of f(x) + f'(x) exists as x goes to infinity and is a finite number (as you said 0:12) , doesn't that necessesarily mean that both the limit of f(x) as x goes to infinity and that of f'(x) exist and that each of them is a finite number ? That would imply that f has some sort of asymptote as x goes to infinity, and that leads us to the conclusion that f'(x) has to go to zero no matter what ... so no need for the e^x condition ... right ????😅 btw I reaaaaaaally love your channel and your content

[@drpeyam]

In theory f and f’ could cancel out! I can’t think of a counterexample right now but think of some wild oscillatory function

[@dbmalesani]

I think you are correct. The proof goes exactly the same way as for the case shown in the video - the hypothesis that lim eˣ·f(x) = +∞ is not necessary, as de l'Hôpital's rule does not require the numerator to be infinite - only the denominator.

[@mario_veca]

@@drpeyam I don't think they can cancel out some crazy behaviour , because if you suppose they do not have two limits going to 0 separately, you have that the hypotesis of the exiting limit for the sum becomes a vinculum for the function, namely f(x) satisfies for large x the differential equation f(x) + f'(x) = L, and this leads to exponential decay for both f(x) and f'(x), countrary to the hypotesis that the two limits vanish separately.

[@naturemeets]

C♥♥L

[@picnicbros]

I just got 100 for Linear Algebra, thanks to you Dr. Peyam! Before discovering you, I find it to be a hard and boring subject. I was told to just use formulas in the textbook and solve problems. I couldnt understand a single thing! But you helped me a lot with your videos. You explained beautifully how and why formulas were that way, and why each concept was useful. Without you, I would have been completely lost. Thank you!

[@drpeyam]

Woooow congrats!!!

[@tgx3529]

I can only imagine the cases where, limit f(x)+f'(x) exists and is finite & lim f(x) exists and is finite, so the function f(x) is bounded, so will be lim f'(x)=0 automaticaly with out suppose lim exp x * f(x)=infinity

[@tgx3529]

@VeryEvilPettingZoo But I suppose that lim f(x)= L (is from R ) for x go till infinity

[@tgx3529]

@VeryEvilPettingZoo Yes, I agree, but there is the condition, that lim f(x)+f'(x) exists, also f'(x) exists, the case limit infinity - infinity is not posible, the case that lim f(x) exist & lim f'(x) doesnt exist is not posible. If both of them limit are finite, then lim f(x)+f'(x) is finite . I think , the derivative f'(x) go to zero, (there is limit from derivative). Or do you know same example, where is it no zero?

[@dbmalesani]

@@tgx3529 I think you are correct. If lim[f(x)+f'(x)) = ℓ (finite) for x → +∞, that is enough to ensure that lim f(x) = ℓ and lim f'(x) = 0 for x → +∞. I wrote a comment yesterday explaining why (sorry for self-referencing, just to avoid repeating).

[@naturemeets]

So C♥♥L . Dr. Peyam.

[@naturemeets]

Thanks to you and it is not as much as the L'hopital rule but as much as for saving the matrix from divergency. You show how to think and Mr Blackredpen shows how to do. the two souls of reason. Like that.

[@l_szabi]

Can we say in general that if a continuous function f(x) has a limit as x goes to inf then lim f'(x) =0 as x goes to inf?

[@drpeyam]

Not true in general, I think there are counterexamples with sin(x^2)/sqrt(x) or something, basically a really wildly oscillating function that goes to 0

[@dbmalesani]

If I understand correctly the question, I'm afraid not. A "classic" counterexample is f(x) = [sin(x²)]/x. By the squeeze theorem, since |[sin(x²)]/x| ≤ 1/x → 0 as x → +∞, we have lim f(x) = 0 for x → +∞. However, f'(x) = [2x·cos(x²)·x - x·sin(x²)]/x² = 2·cos(x²) - [sin(x²)]/x. As you can see, the second term goes to 0 (same argument as before), but the first term keeps oscillating around ±1; for example, by considering the sequence x₁₂(k) = √(2k𝜋) with k integer, lim f'(x₁(k)) = +1 for k → +∞, while if x₂(k) = √[(𝜋+2k𝜋)x], the lim f'(x₂(k)) = -1 for k → +∞. This shows that lim f'(x) does not exist for x → +∞. However, if it is known that lim f'(x) exists for x → +∞, then it must be 0.

[@dbmalesani]

@@drpeyam, I see I gave the same counterexample. I had not seen your comment when I posted mine - I did not intend to duplicate yours! Anyway, great that we converged to the same direction :)

[@l_szabi]

Thanks for your answers, I didn't think of those nasty oscillations.

[@muhammadnuralamsyah5601]

How do you know that lim f(x) exist?

[@drpeyam]

We show it is 0, therefore it exists

[@stevenwilson5556]

@@drpeyam you showed lim f ' (x) = 0, but you didn't show that lim f(x) = 0 or that it exists, and you used lim f (x) in the proof of lim f ' (x) = 0.