Simplification: There is no need to compare the denominator of the sum of both square roots bc all that fraction is strictly less than the numerator given the fact that it is strictly positive. So in order to find the delta(epsilon) you just need to ignore the denominator when continuing with the inequalities. The simplified answer is given epsilon > 0, take delta < min{1, epsilon/5}, etc.
I did a cheat that I think is valid to simplify the denominator: the smallest that denominator can conceivably get (and therefore the biggest the entire expression can get) is if x= 0, so I swapped that denominator out for "1 + sqrt(5)". Your way, though, is more mathematically rigorous, and it better demonstrates the process. I mean, if we're holding that |x - 2| < 1, then x is never going to be 0, so my way is a little bit mathematically inconsistent. But I don't think it's mathematically inconsistent in any way that violates the epsilon-delta method.
@@papabiraneseye You're literally the first person out of the thousands of viewers that have complimented my work and encourage me to keep pursuing it at the same time. I appreciate the love!
Well note that the definition requires that for all epsilon, there exist a delta that satisfies the condition. If you use delta to be Sqrt(x^2+1)+√5 , you might get different values of delta that might not satisfy the condition since the variable of x is changing. That's why we need to restrict the values of x in order to determine the value of delta.
@@helloitsme7553 Well, don't forget that that the existence of the limit doesn't guarantee continuity (it might not be continuous at the point but still have a limit) so it goes beyond that. But yes, you can't use the conclusion to prove the hypothesis.
To everyone, I did look at his other video, and it was really good. A well explained lesson, from someone who wants to share knowledge. Decent fellow! Great teacher!
Simplification: There is no need to compare the denominator of the sum of both square roots bc all that fraction is strictly less than the numerator given the fact that it is
strictly positive. So in order to find the delta(epsilon) you just need to ignore the denominator when continuing with the inequalities. The simplified answer is given epsilon > 0, take delta < min{1, epsilon/5}, etc.
I like it, thank you Alan
Thank you for tackling the problem step by step! Gracias, amigo
No problem, thank you for watching!
sir you just saved my life
I'm happy to hear that sir!
Thank you so much! This was super helpful!
I'm happy!
I like you motive(behind you)..
🙏🙏🙏
Great lecture. You covered a problem I had with at the 16min mark. Thank you for this.
I'm super happy that you were able to get value from it
I did a cheat that I think is valid to simplify the denominator: the smallest that denominator can conceivably get (and therefore the biggest the entire expression can get) is if x= 0, so I swapped that denominator out for "1 + sqrt(5)". Your way, though, is more mathematically rigorous, and it better demonstrates the process.
I mean, if we're holding that |x - 2| < 1, then x is never going to be 0, so my way is a little bit mathematically inconsistent. But I don't think it's mathematically inconsistent in any way that violates the epsilon-delta method.
Nice!
dude you actually might have saved my freshman college life and my overall future. Huge thanks man, really appreciate it!
That's awesome news!!
Great Job!
youre a genius, thank you so so so much
great work
Marvelous
Thank you for your patience 🙏 this is amazing
I appreciate the love
Just wondering why you haven't considered a negative value in the inequality when doing sqrt(x^2+1)?
I like words you put behind as your help
Wow...such elucidation.
God bless you Sir.
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This was really helpful thank you for making it easy for me ❤️
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thank u so much ....☺
learned a lot
Thank you Mehak
Dude your teaching method is the best, pls make a video on continuity, and never stop this noble task
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Thank you
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you more look like like a bro but its more better to call as dear sir. Sir indeed u r amazing ,,,
Hahaha, thank you sir, you too. I greatly appreciate the compliment. Hope is was worth it. Have a good one.
Great explanation !
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Thanks allot very useful ;)
Thanks a lot
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I learned how reciprocals work, I always had a problem understanding how they come from thin area, it because they equal to one.. thanks
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great job
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Brilliant
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Good
Thank you
Great explanation man, I like the way you teach. Please post more stuff on epsilon-delta proofs!
Thank you for the love man, trying my best to help all of you.
Hey Clumpy, thank you for the praise, what other stuff would you like me to post?!
Sir..Can you use the lower bound ??
When we use min=1/2 ?
Nice explanations :-)
Thank you sir
Your screen writing game is on point
Thankyou
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la bestia josh
Aprendiendo de los mejores como tu hermanito
Amazing 💪❤️
Your voice= Obama's voice, I really like it
Haha thank you!
Very good!
Thank you sir
You are welcome! You video is helpful for teacher. I will continue to follow you!
Thank you so much! I'm just trying my best to support free education for all
I strongly encourage you to continue this noble task. Respect!!!
@@papabiraneseye You're literally the first person out of the thousands of viewers that have complimented my work and encourage me to keep pursuing it at the same time. I appreciate the love!
Gracias señorito.
De nada
hey you're great!! may I ask you a problem i need to solve?
Yes Claudia, how can I help you?
Thankyou I actually had an assignment for yesterday! but I will definitely ask if I have more problems!! I really appreciate it :)
Thank you dude , You will get more subscribes !!
Thank you!!
to solve quadratic epsilon delta proof
The moment when he says and now comes the tricky part 18:10
Sqrt(x^2+1)+√5 is all positive therefor you could have written this into delta am I wrong
Well note that the definition requires that for all epsilon, there exist a delta that satisfies the condition. If you use delta to be Sqrt(x^2+1)+√5 , you might get different values of delta that might not satisfy the condition since the variable of x is changing. That's why we need to restrict the values of x in order to determine the value of delta.
@@MOTIVAO oh yeah of course lmao it's like you need to prove that the function is continuous which requires another epsilon delta proof
@@helloitsme7553 Well, don't forget that that the existence of the limit doesn't guarantee continuity (it might not be continuous at the point but still have a limit) so it goes beyond that. But yes, you can't use the conclusion to prove the hypothesis.
The problem is we have to try again delta = (1,epsilon(blabla)) to proof the existence of the limit
Are you referring plug it back in?
how to plug it in and prove that it exists
يعطيك الصحة .....بصح تشبه ولد جارنا
Hi Boubaker, thank you for the comment. I don't speak or read arabic but I appreciate the love.
😂😂 yakhaah
great algebra lesson, but I was looking for someone to EXPLAIN the delta\epsilon limit. Still has no meaning. Thanks anyway.
Check my other video: ua-cam.com/video/9BCUjMaAffM/v-deo.html
To everyone, I did look at his other video, and it was really good. A well explained lesson, from someone who wants to share knowledge. Decent fellow! Great teacher!
Thank you sir, I greatly appreciate the love!
be very quick
Your method is wrong, delta value which you taking is not suitable
Thank you Riki, do you mind elaborating what exactly are you referring to?
@@MOTIVAO yes, i want to collaborating with you, i want to sharing my little knowledge about limit definition together to you
@@RikiFaridoke I have a Instagram: @motivao.math, feel free to send me your knowledge there
"A bit" confusing. Alright lemme just drop out
No doubt this topic is one of the hardest to master but not impossible, don't give up!
@@MOTIVAO I was being sarcastic. I doing well in the class
@@brandonmtb3767 That's awesome
dont c what is the use of this shit and why it is pushed in math, what is it used for?
youre a genius, thank you so so so much
I wish I was one, thank you for the support!