Limits requiring L'Hopital's Rule

[@icafe36485]

How well-mannered you are and how patiently you teach❤

[@PrimeNewtons]

I appreciate that!

[@komalshah1535]

This person is outstanding teacher. So much enthusiasm for maths is unheard of .

[@calebo8882]

just stumbled on your account while studying for my midterm and man you're such a help! i saw one video explaining limits approaching infinity and it was EXACTLY what i needed at that instant. i was stuck somewhere else and searched "prime newtons l hopital" and this video came up and now i understand l hopitals rule kind of. thank you so much! and i hope your day is better knowing that youve helped a random stranger across the world.

[@tomtomspa]

the reason you can bring the limit at exponent is not because e is constant, but rather that the exp function is continuos.

[@spacetimemalleable7718]

Your are absolutely correct.

[@ianrobinson8518]

Very clear demo. And unexpected answer! I would have eased the approach by setting the expression to y and taking logs before applying l’Hôpitals rule. Then exponentiate once -1/2 is derived. It’s also not necessary in practice to keep writing the limit or derivative expression on top and bottom each time. Just go direct. But I accept you’re seeking clarity for demo purposes.

[@codeescape9]

00:08 Limits requiring L'Hopital's Rule 01:55 Applying L'Hopital's Rule to find the limit of the given expression. 03:40 To evaluate the limit, we use L'Hopital's Rule. 05:56 L'Hopital's Rule can be applied to rational expressions if the limit gives zero over zero. 07:53 L'Hopital's Rule allows you to differentiate the top and bottom separately when you have a rational function with a limit of zero over zero. 09:47 L'Hopital's Rule allows us to find the limit by differentiating the numerator and denominator separately. 11:54 The application of L'Hopital's Rule is discussed in finding the limit as x approaches zero. 14:23 The limit of negative secant squared x as x approaches zero is e to the negative 1/2.

[@joshandseb]

"that's a terrible line but we'll still accept it" LOL Your ability to make learning these subjects enjoyable is truly commendable. I want to express my sincere gratitude for your outstanding videos, which are incredibly clear and concise. Whenever I discover that you've covered a topic I'm interested in, it brings me great anticipation and excitement. Thank you so much for your valuable contributions.

[@PrimeNewtons]

🤣 I appreciate the support 🙏 ❤️

[@rudorwashemuzangaza1978]

Sir,you are just too good!!!much blessings!

[@androo_hill]

i loved this video so much, maths is amazing. keep up the enthusiasm

[@GiftMlinde]

SIR I love you so much, u teach veryy well and u make me happy when u smiles oooo GOD bress u so muchhh!!

[@PrimeNewtons]

Amen

[@dimakatsolanga9606]

hilarious😂...good job sir🙌

[@MziweneleDiko]

You're the best sir you deserve an award 👏

[@chaddest]

There is a much simpler method for such questions. Basically, if the limit is of 1^infinity form, you can write as e^[(base - 1) * power]. After doing that, we can write cosx - 1 as x^2/2 as per the expansion series of cosx. And we instantly get the answer. However, i do acknowledge your solution as well because it's just not about finding the solution, it's also about exploring various methods. Thank you.

[@keithrobinson2941]

Those are very large ees! Love the video. Your eeeexplanations areee aweeesomeee!

[@klementhajrullaj1222]

You can do it even more briefly, without using the second derivative, because the limit when x goes to 0 of tanx/x is the same as the limit when x goes to 0 of sinx/x, so equal to 1, therefore, all the limit will be: e^(-1/2)=1/e^(1/2)=1/Ve=Ve/e.

[@nabilahmed7534]

Yeah

[@WisdomMudenda-w3e]

Wow you really know how to explain

[@samsonsisay4762]

You're just awesome!

[@narjsseettegrini]

I love you my brother . And l like your study . Sr

[@KaivalyaChess]

sir u r the best

[@ahmedabdelkoui3790]

From {lim(sinx/cosx)}/{lim(2x)}=lim{sinx/(2xcosx)}=lim{(sinx/x)(1/2cosx)=1/2 because lim(sinx/x )=1. We clearly understand lim in this limiting case when x tends towards zero. Consequently, without going to the tangent and the secant, the limit sought is e^(-1/2)=1/√e.

[@tarciso21claudia28]

Optimus magister in hoc mundo !!!

[@ЕвгенияЛысенко-у4н]

Isn’t it also for infinity over infinity?

[@L1merencer]

Hey i have a question. If i replace the e and natural log to another constant and another logarithm, does the answer change?

[@redpepper74]

It wouldn’t because when you differentiate log_n(x) you get 1/x divided by the constant ln(n). Then, because that constant is in the exponent, you can get rid of it by using the fact that n^(1/ln n) = n^(log_n(e)) = e.

[@bogusawsroda3747]

Super

[@바르고고운말]

L'hopital is very powerful. but you don't need to use that here. there is another solution. in korea, we learn this problem like below. let cosx-1=t, then cosx=t+1, and when x->0, then t->0 (lncosx)/(x^2) =(cosx-1)(lncosx)/(cosx-1)(x^2) ={(cosx-1)/(x^2)}×{ln(t+1)/t}=-1/2 (because lim[x->0](cosx-1)/(x^2)=-1/2, lim[t->0]ln(1+t)/t=1) i am sorry i am not good at english😢

[@nxkluk]

Can't we just add +1-1 to cos(x), and later on substitute t=1/(x^2) ?

[@ACBis]

What If we use log instead ln?

[@ISAACMAJEME]

So far so good

[@surendrakverma555]

Thanks Sir

[@cherryisripe3165]

It’s a beauty!

[@mark23705]

We can also use the fact that sinx/x approaches 1 instead of using the L'Hopital's rule for a second time

[@BinaHejazi]

didn't we say that ONE to ANY (repeating ANY) power is ONE?

[@VenkataPradeep24]

Isn't limit of tanx/x is already 1, If yes, why extra step?

[@GiftMlinde]

❤❤❤❤❤❤❤❤❤!!!

[@jellospaghetti3254]

i love you man

[@spaceshipastro]

thank u!! it went easy

[@mr.mxyzptlks8391]

Not so simple a question. L’Hospital’s rule is from the 16 hundreds. Complex numbers came about in the 17 hundreds. Can L’Hospital apply to C? I never learned about any of this beyond R^1, but I think it probably applies to R^n, comments invited. I don’t think it can be applied to C though. More comments, or references, or proofs, invited 🤓

[@Makunohero]

Cool

[@Emma-ki3fv]

thank you!

[@NStripleseven]

Weird how you started with cos and an exponent and you ended up with e somehow.

[@wafalsh8131]

Can we solve without L Hopital's rule ?

[@subbaraooruganti]

Limit of tan x/x = 1 if x tends to zero

[@林月珍-h2p]

12:12 (lim)(x→0) sin⁡x/x=1

[@Prphadgaming77]

Tanx/x = 1

[@SmokalotOPott]

That is the limit when X approaches 0 from the negative numbers, when X approaches 0 from the positive numbers, the limit is 1.

[@brendanward2991]

But the function is symmetric about the y-axis. Replacing x with negative-x does not change it.

[@NuZhamilya]

its still help me to understand math! Thank you!🫰🏻