In this video showed how and when to use L'Hopital's rule ffor taking limits
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@icafe364857 ай бұрын
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@PrimeNewtons7 ай бұрын
I appreciate that!
@tomtomspa8 ай бұрын
the reason you can bring the limit at exponent is not because e is constant, but rather that the exp function is continuos.
@spacetimemalleable7718Ай бұрын
Your are absolutely correct.
@komalshah153510 ай бұрын
This person is outstanding teacher. So much enthusiasm for maths is unheard of .
@calebo88822 жыл бұрын
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.
@ianrobinson85186 ай бұрын
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.
@codeescape910 ай бұрын
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 Жыл бұрын
"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 Жыл бұрын
🤣 I appreciate the support 🙏 ❤️
@chaddest13 күн бұрын
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.
@rudorwashemuzangaza19784 ай бұрын
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@user-wl4nd7ic1s26 күн бұрын
SIR I love you so much, u teach veryy well and u make me happy when u smiles oooo GOD bress u so muchhh!!
@PrimeNewtons26 күн бұрын
Amen
@klementhajrullaj12224 ай бұрын
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.
@nabilahmed75344 ай бұрын
Yeah
@dimakatsolanga96062 жыл бұрын
hilarious😂...good job sir🙌
@MziweneleDiko3 ай бұрын
You're the best sir you deserve an award 👏
@keithrobinson29418 ай бұрын
Those are very large ees! Love the video. Your eeeexplanations areee aweeesomeee!
@ahmedabdelkoui37904 ай бұрын
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.
@samsonsisay476211 ай бұрын
You're just awesome!
@user-kl6dr9wm3e5 ай бұрын
Wow you really know how to explain
@billykim71796 ай бұрын
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😢
@user-fp7lb4nw9h7 ай бұрын
I love you my brother . And l like your study . Sr
@tarciso21claudia287 ай бұрын
Optimus magister in hoc mundo !!!
@user-wl4nd7ic1sАй бұрын
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@surendrakverma5555 ай бұрын
Thanks Sir
@bogusawsroda37477 ай бұрын
Super
@cherryisripe31657 ай бұрын
It’s a beauty!
@spaceshipastro4 ай бұрын
thank u!! it went easy
@Makunohero5 ай бұрын
Cool
@Emma-ki3fv8 ай бұрын
thank you!
@BinaHejazi10 күн бұрын
didn't we say that ONE to ANY (repeating ANY) power is ONE?
@NStriplesevenАй бұрын
Weird how you started with cos and an exponent and you ended up with e somehow.
@subbaraooruganti2 ай бұрын
Limit of tan x/x = 1 if x tends to zero
@L1merencer8 ай бұрын
Hey i have a question. If i replace the e and natural log to another constant and another logarithm, does the answer change?
@redpepper746 ай бұрын
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.
@ACBis2 ай бұрын
What If we use log instead ln?
@VenkataPradeep245 ай бұрын
Isn't limit of tanx/x is already 1, If yes, why extra step?
@wafalsh81314 ай бұрын
Can we solve without L Hopital's rule ?
@Prphadgaming777 ай бұрын
Tanx/x = 1
@user-dz1du8uk8s7 ай бұрын
12:12 (lim)(x→0) sinx/x=1
@SmokalotOPott8 ай бұрын
That is the limit when X approaches 0 from the negative numbers, when X approaches 0 from the positive numbers, the limit is 1.
@brendanward29916 ай бұрын
But the function is symmetric about the y-axis. Replacing x with negative-x does not change it.