To some certain point, we'll be fluent enough in simple procedures to recognize the patterns, and apply the rule immediately. Simple computational algebra would be very heavy had it not been memorization of special, recurring results
@@Bedoroski Your rationale is correct but that's the wrong approach to learning mathematics. You should always first learn to derive that formula and why that works and not just memorize something a teacher wrote on the blackboard.
@@joaomane4831 I said that in my comment. Work out the problem to the point of fluency, and when you get to a certain level it would be safe to accept key results. I never intend to skip through foundational stages.
Good Morning, Sir, I have done this before but the probem was differed a little : limit of x to -1 (1+x)÷(1+x³) However, they are the same Thank you james 06-21-2024
Hey. I've been trying to learn modular arithmetic. Could you do an introductory video on the topic? It would help so much. I use your videos mainly when studying, because I really resonate with your teaching style.
i just wanna ask something can't we derive the general form using binomial theorem cause x^3+y^3 factors is derived from (x+y)^3 formula and instead of 3 by putting n can 1 instead of y we should arrive at this formula
I did the substitution h=-(1+x^13) then manipulated the limit to equal the derivative of x^(9/13) evaluated at 1 [Limit form]. I am not sure if that's allowed but it's not L'Hopital's rule.
I thought about what if x -> 1, instead of -1. I know that the limit will be 2/2=1, but all manipulations you’ve done are still work, right? (x+1) will be canceled with itself and we are also getting 9/13, right? xddd
Yeah, the 1's will cancel with the −1's, so then we can factor out an 𝑥 in the numerator and denominator. 𝑥 ∕ 𝑥 = 1 and all the other 𝑥-terms evaluate to 0, leaving us with ₉𝐶₈ ∕ ₁₃𝐶₁₂ = 9 ∕ 13. Neat!
@@sohaib_mer- Because of the binomial theorem. When expanding (𝑥 − 1)⁹ we get a 9th degree polynomial in 𝑥 whose coefficients can be determined through combinatorics.
@@jumpman8282 thanks man, the thought of using binomial expansion theorem at all, and had just finished learning it(2 weeks prior to this convo), hope you're having a good day and again, many thanks
If you did not know about this approach, then you could just "guess" that x + 1 divides the numerator and the denominator. Why this guess? well because if this is true, then the troublesome factor of x + 1, with limit x + 1 = 0 as x -> -1, would cancel in the denominator and the numerator and the limit of the remaining expression could be computed 😂.
Before the species was found, people were: Edit: When we do lopital in this section, we get 9x^8/13x^12 when we find a separate derivative. Since x is close to "-1", we should put "-1" where we see x.
Later in calculus, you will approach some limits where using l’ hopitals rule actually doesn’t work out for evaluating the limit. And besides, not using this hopitals rule is a great way to work on your problem-solving skills 😊
first of all i have never worked with the lim() funktion! i am just going to a comersial school in germany but ok i will try to follow you like before 1:05 that law i hopfuly don´t need in future because i forgot it at the moment you finished to say it :D to answer the question: i think i would try to make an equantion and reform it but i have no idea how it could looks like! 1:25 i have to repeat that i like the intro (funfackt the acutal minut here withaut ":" is 5³ ) 2:11 ok nice information 3:55 but why you don´t use the product sybol here? 7:19 the simpelst multiplication ever if you take the sum and make a multiplication 8:08 ok i was right you have to reform it and finde a better and easyer way nice! 8:18 ok if i translated it right it aktualy make sence! see you K.Furry
Instead of memorizing that factorizing formula, just know how to employ synthetic division. It's that simple.
To some certain point, we'll be fluent enough in simple procedures to recognize the patterns, and apply the rule immediately. Simple computational algebra would be very heavy had it not been memorization of special, recurring results
I also found it using synthetic division
@@Bedoroski Your rationale is correct but that's the wrong approach to learning mathematics. You should always first learn to derive that formula and why that works and not just memorize something a teacher wrote on the blackboard.
@@joaomane4831 I said that in my comment. Work out the problem to the point of fluency, and when you get to a certain level it would be safe to accept key results. I never intend to skip through foundational stages.
Me: I solved it without knowing what this "L'hospital" rule is 😂
Hello sir I have seen your video of tetration and I can't understand what you did.The ques was 2 tetrated 3.Please reply me.😊🎉❤
why not just use lhopitals rule. 0/0.?
Nice explanation. I am a fan of your teaching!
Which software are you using to write mathematical expressions on your thumbnail?
PowerPoint
@@PrimeNewtons thanks for response😊👍 waiting for your next interesting problem😊
Good Morning, Sir, I have done this before but the probem was differed a little : limit of x to -1 (1+x)÷(1+x³)
However, they are the same
Thank you james 06-21-2024
lim x-> -1 1+x^9/1+x^13
lim x-> -1 -1(-1-x^8)/-1(-1-x^12)
lim x-> -1 -1^2(-1-x^6)/-1^6(-1-x^6)
lim x-> -1 -1^2/-1^6
lim x-> -1 -1^2/-1^6=1/1
lim x-> - 1 =1
Outstanding, you bet. Regard’s from Brazil.
long division worked for me
Hey. I've been trying to learn modular arithmetic. Could you do an introductory video on the topic? It would help so much. I use your videos mainly when studying, because I really resonate with your teaching style.
I'll consider it
@@PrimeNewtons Thank you.
in general, for odd m,n, we have the limit x --> -1 (1 + x^m)/(1 + x^n) = m/n
This is great man, always loved your channel since day 1. Thank you for the videos.
Thanks for the support 😊
i just wanna ask something can't we derive the general form using binomial theorem cause x^3+y^3 factors is derived from (x+y)^3 formula and instead of 3 by putting n can 1 instead of y we should arrive at this formula
That was great! Thanks so much! 👍
I did the substitution h=-(1+x^13) then manipulated the limit to equal the derivative of x^(9/13) evaluated at 1 [Limit form]. I am not sure if that's allowed but it's not L'Hopital's rule.
Very nice!
I thought about what if x -> 1, instead of -1. I know that the limit will be 2/2=1, but all manipulations you’ve done are still work, right? (x+1) will be canceled with itself and we are also getting 9/13, right? xddd
One can also substitute x -> x-1 and limit x -> 0, then raise to the power
Shush, let the master add up the ones.
Yeah, the 1's will cancel with the −1's, so then we can factor out an 𝑥 in the numerator and denominator.
𝑥 ∕ 𝑥 = 1 and all the other 𝑥-terms evaluate to 0, leaving us with ₉𝐶₈ ∕ ₁₃𝐶₁₂ = 9 ∕ 13. Neat!
@@jumpman8282 hey!, can i ask you how/why u used combinatorics to get 9/13
@@sohaib_mer- Because of the binomial theorem.
When expanding (𝑥 − 1)⁹ we get a 9th degree polynomial in 𝑥 whose coefficients can be determined through combinatorics.
@@jumpman8282 thanks man, the thought of using binomial expansion theorem at all, and had just finished learning it(2 weeks prior to this convo), hope you're having a good day and again, many thanks
Excellent video
l'hospital rule is my friend..... how come it is now allowed....
We want use hopital rule master... thanks for an other video.
Very good. Thanks 🙏
What a fantástic teacher!
If you did not know about this approach, then you could just "guess" that x + 1 divides the numerator and the denominator.
Why this guess? well because if this is true, then the troublesome factor of x + 1, with limit x + 1 = 0 as x -> -1, would cancel in the denominator and the numerator and the limit of the remaining expression could be computed 😂.
It's not a guess that x+1 divides both. It's a clear fact, given that -1 is a clear root of both.
@@xinpingdonohoe3978
That's why it is a good guess 🙂
It is a simple application of Ruffini's theoremstuuf usually learned if first high school year.
@@sauzerfenicedinanto
Think that you are 13 and smart. No theorems required 🤣😂.
I would have enjoyed watching you use l'Hopital's Rule to confirm the answer you got algebraically.
easy af tbh
Before the species was found, people were:
Edit: When we do lopital in this section, we get 9x^8/13x^12 when we find a separate derivative. Since x is close to "-1", we should put "-1" where we see x.
x^{2n}+a^{2n} , can be factored into quadratic factors
Not for 𝑛 ∈ ℤ.
𝑥^(2𝑛) − 𝑎^(2𝑛), though.
Great❤
Whats the point in solving without lhopital??
Later in calculus, you will approach some limits where using l’ hopitals rule actually doesn’t work out for evaluating the limit. And besides, not using this hopitals rule is a great way to work on your problem-solving skills 😊
@@MathProdigy-qg5gx I have already completed calculus
L'Hopital is like a cheat code for limits. If you can solve complex limits without resorting to that rule, then you're golden at Algebra.
first of all i have never worked with the lim() funktion! i am just going to a comersial school in germany but ok i will try to follow you like before
1:05 that law i hopfuly don´t need in future because i forgot it at the moment you finished to say it :D
to answer the question: i think i would try to make an equantion and reform it but i have no idea how it could looks like!
1:25 i have to repeat that i like the intro (funfackt the acutal minut here withaut ":" is 5³ )
2:11 ok nice information
3:55 but why you don´t use the product sybol here?
7:19 the simpelst multiplication ever if you take the sum and make a multiplication
8:08 ok i was right you have to reform it and finde a better and easyer way nice!
8:18 ok if i translated it right it aktualy make sence!
see you
K.Furry
I believe that you have a problem with L'Hospital him self😂
L'Hopital's rule is for those who think the point of a journey is to arrive.