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.
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.
"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.
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.
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.
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.
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.
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.
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 🤓
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😢
![Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]](http://i.ytimg.com/vi/nJ9lEolfVpU/mqdefault.jpg)
How well-mannered you are and how patiently you teach❤
I appreciate that!
This person is outstanding teacher. So much enthusiasm for maths is unheard of .
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.
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.
the reason you can bring the limit at exponent is not because e is constant, but rather that the exp function is continuos.
Your are absolutely correct.
Sir,you are just too good!!!much blessings!
"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.
🤣 I appreciate the support 🙏 ❤️
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.
i loved this video so much, maths is amazing. keep up the enthusiasm
SIR I love you so much, u teach veryy well and u make me happy when u smiles oooo GOD bress u so muchhh!!
Amen
hilarious😂...good job sir🙌
You're the best sir you deserve an award 👏
Those are very large ees! Love the video. Your eeeexplanations areee aweeesomeee!
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.
You're just awesome!
sir u r the best
Wow you really know how to explain
I love you my brother . And l like your study . Sr
Hey i have a question. If i replace the e and natural log to another constant and another logarithm, does the answer change?
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.
Super
Isn’t it also for infinity over infinity?
So far so good
Thanks Sir
i love you man
It’s a beauty!
Can't we just add +1-1 to cos(x), and later on substitute t=1/(x^2) ?
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.
Yeah
thank u!! it went easy
What If we use log instead ln?
Optimus magister in hoc mundo !!!
Isn't limit of tanx/x is already 1, If yes, why extra step?
Cool
thank you!
didn't we say that ONE to ANY (repeating ANY) power is ONE?
❤❤❤❤❤❤❤❤❤!!!
We can also use the fact that sinx/x approaches 1 instead of using the L'Hopital's rule for a second time
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.
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 🤓
Weird how you started with cos and an exponent and you ended up with e somehow.
Can we solve without L Hopital's rule ?
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😢
Limit of tan x/x = 1 if x tends to zero
12:12 (lim)(x→0) sinx/x=1
Tanx/x = 1
That is the limit when X approaches 0 from the negative numbers, when X approaches 0 from the positive numbers, the limit is 1.
But the function is symmetric about the y-axis. Replacing x with negative-x does not change it.
its still help me to understand math! Thank you!🫰🏻