My professor prooved the same thing to us. He skiped all the (un)necessary comments what's coming from where so I was so lost looking back at my notes. Your videos are for sure saving me :D
First I need to tell you that you have great videos with excellent explanations. These definitely make math learning interesting, enjoyable and easy to understand. The point of this comment is to let you know about a very important notation matter. When we right limit at the beginning without a parenthesis, the limit only belongs to the first term. So, the correct way to write it is to use a parenthesis around all three terms as lim{ x → 1} (x²+5x+6) = 12. Keep doing what you are doing. Thank you.
This is the best I have ever heard this concept explained. I was struggling with understanding how the proof worked for the past 2 days, and I finally understand now. Thank you!
Hello Sir, there is a mistake in this proof. You cannot replace |x+6| with 8 and be sure that 8|x-1|< E. We can take 6 instead of 8 but in that case in the last step you wont be able to replace |x+6| with 6, hence we get stuck. Could you please check for this.
Hey, first of all, i really like your videos. You are doing a great job! I am a fan of generalizing the epsilon-delta-criterium and give a specific formula for delta. What I found out was that, if you choose for a general x_0 (which is in this case 1 of course) the formula delta (epsilon) = - abs(x_o) - 5/2 + sqrt( epsilon + (abs(x_0) + 5/2)^2), you get a guaranteed delta for any epsilon > 0 that you choose. Thank you for your great videos again. They are very inspiring!
Can someone please explain when 9:04 he writes x+6 as 8 in the inequality. Because he showed that x+6 is less than 8, but when he inserts 8 in the epsilon inequality, he technically increased the left hand side of the inequality without increasing the right hand side (epsilon). Which makes me a little suspect on whether the inequality will hold true. I am no professor and I respect everything he does, but I get delta=epsilon/6. Please feel free to enlighten me. Cheers from Denmark
You are really doing an excellent explanation and thus make our life easier in understanding maths even though it is not my proffesion. Awesome! Thank you!
this is the serious issue what am i going to tell you.... if 6 < x+6 < 8 and at the same time |x+6|.|x-1| 6 and now replace x+6 with 6 then it will be not contradict because the product of |x+6|.|x-1|
When you want to show that the left hand side is less than the right hand side, you should always increase the left hand side. So 8 and not 6 was the correct choice.
In your linear example, you had the process right, but you stated it as “for every delta, there exists epsilon.” It is an easy mistake to make. I actually had to read an epsilon-delta proof in my textbook to understand why it had to be “for every epsilon, there exists delta.” We first find the delta and then prove that it implies the given epsilon.
It depends on what your teacher wants. Some just want you to show there is a delta. But some want you to show that that value you found truly works. So ask the professor what they want.
My professor prooved the same thing to us. He skiped all the (un)necessary comments what's coming from where so I was so lost looking back at my notes. Your videos are for sure saving me :D
The explanation is all wrong.
Sir, could you recommend a better video? @@tomtomspa
@@tomtomspait’s not
This is it. Two hours of working on the same delta-epsilon problem, and this is the video that made it click for me. Thank you.
Mathematics becomes easier for me because of how you explained it in such simple ways....Thank you very much
It was wrongly explained!
@@yohannesyebabe7923 I cannot believe how someone so poorly knowledgeable on limits get to wrongly explain them on yb
My first time to appreciate someone on RUclips
Thanks
First I need to tell you that you have great videos with excellent explanations. These definitely make math learning interesting, enjoyable and easy to understand. The point of this comment is to let you know about a very important notation matter. When we right limit at the beginning without a parenthesis, the limit only belongs to the first term. So, the correct way to write it is to use a parenthesis around all three terms as lim{ x → 1} (x²+5x+6) = 12. Keep doing what you are doing. Thank you.
Unfortunately that's the least serious error of the video.
9:18 I did the same expression and the meme appeared! Truly magnificient! YOU GOT ME! LOVE FROM INDIA
You're really making our life much simpler and thanks a lot. You simplified the proofs very well.
I love the way you are so enthusiastic about teaching!! thank you so much for the video :)
The proof has fundamental error!
The correct proof is as follows
Let f(x)=x^2+5x+6
We want to show lim x->1 {(f(x))}=12
For all e>0 whenever |f(x)-12|
Thanks a lot buddy! ✌🏻💚
This comment deserves a lot more likes.
Why would you need this type of math
@@brendawilliams8062 It is the rigorous mathematical analysis called the epsolon delta techniques.
@@yohannesyebabe7923 thx. , but what can you do with it
@@brendawilliams8062 To prove mathematical theorem.
You are One of the Great Teachers i have ever seen Thank a lot!!!!
This is the best I have ever heard this concept explained. I was struggling with understanding how the proof worked for the past 2 days, and I finally understand now. Thank you!
SO AM I FROM University of Namibia 👌👌👌👌👌👌
Because lx+6l
He should replaced by 6.
and
He should choose delta a min {1, e/6}
Thank you so much for the explanation. Watched bunch of other videos and yours was the simplest to understand
Bro you are amazing , i mean you have the talent to be a teacher bc you try to understand the students❤❤
Thank you
A very very brilliant lecture from a genius . Thank you sir for your contribution.
You are the best math teacher that exists!
I actually appreciate you for the good work ,,,God bless you
Your handwriting is dope sir and I love the way you have explained.
Thank you!
J has to pass my Calculus I course knowing that i will fail in every question of this type. Great explanation!!! Thank you
Plus the way u smile doing some cracks, while solving, its just fantastic, u wont be left with any qsn
Hello Sir, there is a mistake in this proof. You cannot replace |x+6| with 8 and be sure that 8|x-1|< E. We can take 6 instead of 8 but in that case in the last step you wont be able to replace |x+6| with 6, hence we get stuck. Could you please check for this.
That's exactly what I also figured out...👍
We may take delta=min{1, epsilon/8}
Hey, first of all, i really like your videos. You are doing a great job!
I am a fan of generalizing the epsilon-delta-criterium and give a specific formula for delta. What I found out was that, if you choose for a general x_0 (which is in this case 1 of course) the formula delta (epsilon) = - abs(x_o) - 5/2 + sqrt( epsilon + (abs(x_0) + 5/2)^2), you get a guaranteed delta for any epsilon > 0 that you choose.
Thank you for your great videos again. They are very inspiring!
I understood this today 😢
You are a great teacher 👏
best video on the subject so far.
Never Stop Teaching!
You are a Geneous sir...Excellent in making mathematics simple ..very good in explaining
I want to be a genius. It's just hard 😪 🤣
Wooow Man u the best, i actually had a crisis in understanding this, but i just grasped it easily, The best of the Bestest
Can someone please explain when 9:04 he writes x+6 as 8 in the inequality. Because he showed that x+6 is less than 8, but when he inserts 8 in the epsilon inequality, he technically increased the left hand side of the inequality without increasing the right hand side (epsilon). Which makes me a little suspect on whether the inequality will hold true. I am no professor and I respect everything he does, but I get delta=epsilon/6.
Please feel free to enlighten me. Cheers from Denmark
You my friend are an excellent tutor. Great job on Precise Definition of a Limit !!! Keep up the good work....sorry... Great Work.
Very helpful master ❤❤❤ from India.
It's Amazing. I watched many videos but after watching this video I understood the math well.
Even i can miss up a class,, its okay with this handful of skills, i really appreciate you.
thank you very much i was really confused on this delta and epsilon proofs you just made my day,thank you sir
Thank you very much sir
❤ From india 🇮🇳
You are really doing an excellent explanation and thus make our life easier in understanding maths even though it is not my proffesion. Awesome! Thank you!
Bro keep uploading...Love from Bangladesh 🇧🇩
Like how you explain it my brother.much appreciated.
Damn , that's some quality teaching .
He does it again! Bam! QED! 😃
Your enthusiam for maths makes me like it
By far best explained thankyou
Glad it helped
your videos are very interesting keep going on !
why x-6 cannot be replaced with 7? it is smaller than 8 how we replaced that?
Very good sir love from antartica
this is the serious issue what am i going to tell you....
if 6 < x+6 < 8 and at the same time |x+6|.|x-1| 6
and now replace x+6 with 6 then it will be not contradict because
the product of |x+6|.|x-1|
i dont know you are looking for this or not.....
i hv figured out the mistake..
you should not start with lx^2 +5x-6l
Yes you are right. I spotted that straight away.
When you want to show that the left hand side is less than the right hand side, you should always increase the left hand side. So 8 and not 6 was the correct choice.
wow superb sir ❤
Truly amazing
thank you so much sir, it really helps me a lot. And now, maybe I can answer our exam tomorrow hhahhaha
Good luck
AWESOME SIR YOU ARE JUST A GREAT MAN THNX A LOT SIR
Great video! Thanks. It would be nice to add some graphics to bring this concept to clarity.
THANK YOU SO SO SO MUCH YOU HELP ME A LOT A REALLY LOVE YOUR EXPLANATION
Love that t-shirt!
You're so talented
I finally understand this brother thanks a lot
really exciting
Bravissimo.
Con te e con Robert Ghrist la matematica è un incanto.
Grazie 😊😊
what a nice video to watch
Can you please try this problem
Use Epsolon delta Def to show that
Lim x-->-1.(2x+3)/(x+2)=1
God bless your soul. Thank you
Amen
Love❤ from India
Ohh you are so so good❤
Thanks help me a lot
Wonderful sir. You are wonderful..
Best love from India❤
In your linear example, you had the process right, but you stated it as “for every delta, there exists epsilon.” It is an easy mistake to make. I actually had to read an epsilon-delta proof in my textbook to understand why it had to be “for every epsilon, there exists delta.” We first find the delta and then prove that it implies the given epsilon.
Oops, that was an error.
8:58
We substituted 8 because we
|x-1
It is good. I am lucky if you add another example.
If not possible thank for your videos
Thank you so much for the very clarification🎉❤
Thanks for video.... epsilon and delta greater than zero is true
Thanks man, I finally got it
Glad it helped
great job Bosses
Thankyou in advance
Great teacher! Thanks...
U're genius
Respect from Cameroon +237, 3:19 🎉🎉🎉
Love the 2+2=4 shirt
Man u save my life
which one is dependent
That is easy and clear , do you do one to one tour ?
Yes. Please send me an email
So helpful sir amazing ❤
You are great
well taught sir
wow u r great man
Thanks so much 🙏
Nice explanation
in the second part can we not substitute delta = epsilon/8 into |(x-1)|< delta for the proof to be satisfied.
It depends on what your teacher wants. Some just want you to show there is a delta. But some want you to show that that value you found truly works. So ask the professor what they want.
@@PrimeNewtons thanks professor.
Do you have calculus 1 playlist?
Countless hours of calculus and epsilon delta videos and I still have no clue what is going on
Great!
thank you so much!
Nice❤
Nice dress sir
What if you are asked to proof lim
X--2 2x² +3 =11 .
How can I go about it?
you"re bestttttttttttttttttttttttttttttttt
I love it 🎉
Every time I see these videos on delta-epsilon I come to the conclusion that the proof is always a tautology 🤷🏻
I speak portuguese but undertand very well
Sir which camera are you using
Here I can't see the value of proving this as a limit , since one can just plug in the value 1 into this continuous function.