Epsilon-Delta Proof of Limit (Quadratic Example 2)
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- Опубліковано 5 жов 2019
- In this video we do an example of using the formal definition of a limit for an epsilon-delta proof of the limit of a quadratic function. There's one tricky part where we assume that delta = 1. After making the assumption we find an upper bound on an expression and work from there. Setting delta equal to 1 is a very common thing to do!
This is not a topic that you need to worry about for the AP Calculus AB or AP Calculus BC exam (definitely true in 2020...and since maybe the 1980s).
This is the most critical point within explanation to this kind of question. Well done.
“ If d=1,|2x+3| |x-3| < 11. Otherwise, d=e/11
Brilliant Educator . U made each bit of it easy to comprehend. Thank u so much.
thanks this is helpful for my self study in calculus
I've been watching a lot of epsilon delta proofs to understand it and this one helped me
Life saver, i have a calculus exam in three days and proving limits always boggled my mind. Thank you so much
Good luck!
helped so very much. thanks from Finland
Amazing video
Thanks!
Amazing video!
👑👑👑👑👑👑you dropped this, King
Take your W
Godd explanation! Thanks!
Thank you so much
Dear Teacher, could you make one example , which is not factorable ,
I don't understand how you can say |2x+3|
Where did you come up with 4+3
It's not equivalent but we establish that the 2x+3 term cannot be more than 11. 11 is the upper bound. So we just substitute in 11 in place of 2x+3.
You're probably thinking but what if 2x+3 is less than 11. Well it may be, but the point of the exercise is to find a delta that gets us epsilon close to the limit. A delta, not necessarily the optimal delta. We are solving for the largest delta that will give us what we want.
the technique used in the video is reverse implication, where the whole proof is in reverse,
so if a+3
but keep in mind the left side is only a draft, formal mathematicians may write “it suffices” in the beginning of every line if they are lazy
...crystal clear!!!....can you do a cubic?....
Excellent explanation,thank you
Glad it was helpful!
MY BIGGEST THANKS GOES TO YOU
thanks my dude, that's some gold.
Clean explanation!
Thanks!
Omg thanks !!!
Yep this is what i needed. Textbooks complicate things
at 4:57 in the video, I wonder about how [ |2x+3||x-3| < epsilon ] became [ 11*|x-3| < epsilon ]. How can we know that epsilon is greater than 11*|x-3|? Cause we only know that |2x+3| < 11 so we can say that [ |2x+3||x-3| < 11*|x-3| ] but it doesn't always imply that [ 11*|x-3| < epsilon ]. I think it can be like [ epsilon < 11*|x-3| ] too.
I used the sentence from vieta to factorise. In this case: p*q=ac, p+q=b, p*q=-18, p+q=-3, 3*(-6)=-18, -6+3=-3 Then comes: 2x^2-6x+3x-9=2x*(x-3)+3*(x-3)=(x-3)*(2x+3)
wonderful.anx
Calc exam tmr
calc final tmr
for the delta = min { 1, epsilon/11 } part i get delta = min (1 / epsilon / -4 } is it ok if it is a negative number
i meant not for this problem but for another quadratic problem my factors were (x-1)(x-6)
why people working on this proof with the easy quation always assume that Delta just equals 1.
(PS: I am just started learning Calculus days ago, so I only see few example questions which is easy because its purpose is to teach the epsilon delta proof)
Is it just to make the whole process more easily to be done and be easily understood?
Otherwise, i am wondering if i can assume randomly that the positive constant Delta equals instead of 1?
I know it is the unnecessary step.
I think you can take any number but 1 is just convenient
You have to know the reason why he chose 1
The limit function he was given had x->3, which means that 1 is actually really close to c, hence can be used. If it was x->50 you're better off using a closer number
@@uwu.-.5873no, the limit function doesn’t matter, even if x went to 50, 1 would still be as valid of a choice, delta is the upper bound of the difference in between a value chosen for x and the point x is going to, which in our case is 3.
What does "=>" mean? I keep seeing it used lately and I just don't know what they mean. Same thing with "
It means that the thing on the left implies the thing on the right. So something like, horse => 4 legs (assuming all horses have four legs). means that both things imply each other, so knowing one of them tells you the other is also true. Hope this helps!
turksvids thank you so much
I now feel mentally fortified to tackle oddball manipulations.
How do you know that E/11
So, I'm not 100% sure I follow your question but as to the last part, yeah, the largest value of delta that I would accept is 1 but in practice since epsilon will be infinitesimally small, epsilon/11 will be smaller than 1.
this is so helpful massive thanks but i still wonder why delta is equal to 1 and not less than 1 is it the same?
he is aking an assumption so that the eq 2x+3
That's great!
But, you have to take delta as epselon/7 not epsilon/11 as you have to take the reciprocals.
Thanks.
absolute value of 7 and 11 means 11 is the one we should choose i think.
how can delta be negative when it's the distance from x?
where is delta negative?
@@turksvids -1
thnks. was helpful
I've been chipping at counter examples of this, where you choose the wrong limit and see what happens to the math. I think your style and knowledge would be excellent for that type of video.
that's a really interesting idea and definitely might help some people to better understand the definition/work. something like "we've found an epsilon for which there is no delta!"
@@turksvids @turksvids from what I can see, rational functions work in this context by the strictly multiplicative nature of the composition of their constituent polynomials. You can convert a rational function f(x)=p(x)/q(x) where deg(p(x))=n and deg(q(x))=m into a form where you have (x -> a) => f(x) -> L, equal to a coefficient times the product of (x-b_i)^(e_i) from i = 1 to i = n+m and where e_i is either 1 or -1. This means that you can do all of your cancellation on the bounds via multiplication. There's some algebra here that I'm missing, but it feels like there are some results from ring/field theory that will link this the existence of (x-a) as a root of f(x) and that if this weren't a strictly multiplicative construction, you couldn't show that 0 is a limit point of delta(epsilon), delta as a function of epsilon.
Why did you pick delta = 1
It's an arbitrary choice. For every epsilon there has to be a delta, so we can say that, sure, if delta is 1 (which is a simple, convenient choice), we stay in the epsilon-neighborhood of the limit when we stay in the delta-neighborhood of c (the x-value where you're evaluating the limit). In practice you'll never be using 1 for delta, you'll always be picking the delta that's a function of epsilon because that will be the minimum of your delta as a function epsilon and 1. Not sure if this is super clear, but that's the idea.
Hi! May I ask how did you say that epsilon/11 is less than 1? So doesn't that mean that delta is just equal to epsilon/11 instead of min{1, epsilon/11}? Hope you could answer this. Thanks!
Confusing lecture
sorry, hope you find something that works for you! link it back in a comment so others can benefit!
overall awesome explanation and super helpful! i think u made a mistake when u removed 2x+3 from the absolute value. u proved that 2x+3 is less than 11, u did not prove that the absolute value of 2x+3 is less than 11. if u had kept the 2x+3 within the absolute value, u could have replaced that with the 11.
2x+3 is greater than 7 implying that it is a positive number hence why it’s completely fine to say |2x+3| is less than 11. It’s already positive so taking an absolute value won’t do anything
I sent you an email. If you can take a look.
Dreadful explanation. You acknowledge what is confusing but fail to provide substantive answers to make these points make sense.
Heh. Thanks for watching? Please share your explanation so others can benefit!