proving the limit of a product is the product of the limits, epsilon-delta definition
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- čas přidán 16. 07. 2024
- We will prove the limit of a product is the product of the limits (assuming the limits exist) by using the epsilon-delta definition of a limit. This is a classic proof that you will see in your advanced calculus class or real analysis class. This is hard proof especially when you first see it. Be sure you put in the effort and study it well. #calculus #realanalysis #blackpenredpen
Notes for this video are available on my Patreon: 👉 / notes-proof-of-82385985
Introduction and more examples of the epsilon-delate definition of a limit: 👉 • Epsilon-Delta definiti...
0:00 Review some limits and their epsilon-delta definition
2:48 The proof!
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This video is nonsense. The proof that the multiplication function is continuous is trivial, and the proof that the composition of continuous functions is continuous is also trivial. (x+epsilon)(y+delta) = x y + epsilon*x + delta*y +epsilon*delta, and all quantities are infinitesimal aside from x*y. This observation can be turned into an epsilon delta proof automatically by rote. It's ridiculous to call this difficult, it's obvious.
@@annaclarafenyo8185 f and g might be discontinuous
@@blackpenredpen Continuous AT THE POINT means the limit of f equals M and the limit of g equals N, you can set the values to be M and N, and then they are continuous, and the continuous composition of continuous functions is continuous. These methods are formalizations of infinitesimal arguments that are obvious: (f+df)(g+dg)=fg + infinitesimal. There is no work in formalizing this. Making this obvious nonsense sound difficult just serves to mystify epsilon-delta, and make it a hazing ritual rather than a method of proof you internalize.
@@annaclarafenyo8185 f and g don’t need to be continuous at a
@@blackpenredpen IF YOU DEFINE THEM SO THAT THE LIMIT IS EQUAL, THEY ARE BY DEFINITION CONTINUOUS AT A. You should not be teaching matheamatics.
You know something is wrong when bprp doesn't use whiteboard for proof
Simply if he doesn’t use a whiteboard (than something has changed)
Actually my teacher showed us this proof in grade 11, I don’t think it’s too hard (now we weren’t supposed to learn the proof but showed us nonetheless)
I needed to keep everything on the board for this video thus I chose the iPad route 😆
@@blackpenredpenwhat app did you use on your ipad? would it be good for notes at university?
@@Paul-ob2hy I never took notes on an iPad when I was a student. However, I have been using the app "Good Notes" and I have been loving it!
27:05 to avoid the 0/0 case, we can simply have epsilon*|L|/[2(|L| + 1)] < epsilon*(|L| + 1)/[2(|L| + 1)] = epsilon/2, and the solution is unaffected in any way ^^
Ahhhh. I didn’t think of that 😆
Or you can observe that 0
Yeah i was thinking he would do that
Not to kiss your behind or whatever, but this is the most helpful channel I’ve ever found when it comes to helping me with my EXTREMELY hard math class
I am happy to hear! Thank you!
This is one of the more complicated proofs of basic analysis facts and I've never really had a good intuition for the quantities that come up in the proof, so thank you for doing a good job of talking through it.
26:07 I saw some people trying to explain how we don't have to bother about the |L|=0 case in a certain way, but in the end :
We know that |L| < [L|+1 for any real number L. So |L|/(|L|+1) < 1 as a consequence. Multiplying each sides by eps/2 (> 0) guarantees us the result !
Edit : Nevermind it has been said by multiple people already, my bad
1:28 This is the best explanation of the epsilon-delta definition of limits I have ever seen. Everything for me just clicked once I saw this. You're an awesome teacher!
Thank you!
This is blackpenredpenbluepengreenpen and yellow highlighter
😆
@@blackpenredpen 😂
26:27 You don't have to cancel two zeros if you add 1 to |L|, proving that step rigorously!
|L| * (epsilon)/(|L|+1) < (|L| + 1) * (epsilon)/(|L|+1) = (epsilon)/2
was going to write this 👍.
Ah yes. I didn’t think of that. Thanks for letting me know.
tzovgo, you made a little mistake in the formula, which I didn't notice at first, neither did, I presume, BPRP. We both understood what you meant, but you forgot to put the 2 in the denominator whenever |L| was present in the quotient. This is what it should be:
epsilon*|L|/[2(|L| + 1)] < epsilon*(|L| + 1)/[2(|L| + 1)] = epsilon/2
extremely good job and explanations. As a new online teacher, I find your content really inspiring. I now recognize that I mimic a lot your teaching style. Keep up the good job !
Thank you for your kind words. Wishing you all the best! : )
I'm graduating soon in Mechanical Engineering and I just have to say your channel is the best! You've helped me so much!
Thank you. : )
Bprb, WHAT AN AWESOME PROOF! Even though it turned out to be a very difficult one, I understood every step of it, and that's because you are en EXCELLENT TEACHER. Thank you! 🙏🏻
I prefer to use epsilon-delta arguments only in the simplest cases. For example,
- A sum of two functions approaching 0 approaches 0
- A function approaching 0 times a bounded function approaches 0
Now f can be expressed as L plus a function approaching 0, and g can be expressed as M plus a function approaching 0. You can just multiply it out and see that the product is LM plus a function approaching 0 i.e. the limit of the product is LM.
I have always loved working out these fundamental proofs. I actually did this particular one out just a few months ago working through my Schaum's Calculus workbook. I didn't think it was particularly difficult- the really key insight is that addition/subtraction trick that allows the breaking apart of the functions, but it is a common technique in algebra, so came to me almost instantly.
Thank you very much!!! I Found this proof in Spivak's book but I was not getting how it was done and after watching your video now is all clear to me. Love you!
Excellent explanation 👍
Wow - this takes me back to my modern analysis course in college. I'll never forget how brutal that was. But good job with the proof!
Thanks, Scott!
I remember getting this question in my university practice and getting the proof myself is so satisfying!
the moment we draw that box at the very end!!
You are looking CLEAN on the thumbnail man. Nice
Main trick is writing g(x) as
g(x) -M+M after this there is no need to worry about the term f(x) g(x). It is all about being careful. We have to adjust things.
Seeing math makes me remembers when i find the answer
Like "I have 9 numbers, the first 5 numbers has a mean of 12, while the last 4 numbers has a mean of 3, what is the mean of all 9 numbers"
And thats when i realise i can just do (5 × 12 + 4 × 3) ÷ 9 and got 8
You could also just prove the rule for limits of sequences, and then use the sequential definition of the limit to get the same result for limits of functions.
Another approach would be to prove the result in the special case L = 0 = M. This is quite simple as you just choose δ such that
|x - a| < δ ⇒ |f(x)| , |g(x)| < √ε
Then for the more general case define new functions:
F = f - L, G = g - M
Now the limit of both F and G is just zero, so the limit of FG is zero. Do a bit of expanding and out comes that the limit of fg is LM.
Thank u so muchhhhhhhhhhh!!!!!!!!! U save me life by this wonderful video, thank uuuuuuuuuuuuuuuuuuuuuuuuuu!!!!!!!!!!!!!!
Amazing video, thank you so much
I find it easier in the general case of a finite dimension vectorial space like R^n and considering a subordinate norm which is sub-multiplicative... Of course you need a bit of topology but the proof is just much nice
I know the one for sequences and it is quite hard. I had to make my own one because I couldn't remember the one from the lecture. Anyway, from that you can easily arrive at the one for functions thanks to the sequence definition of a limit
Flashbacks to Real Analysis I where our professor wanted us to figure out how to do this proof on our own *shudders*
That is why you derive the sequence criterion for limits, and then you can use the easier to prove limit theorems.
I think the easiest way to think about the 0/0 part is: if |L|=0, then |L|*epsilon/(2*(|L|+1))=0 which is definitely smaller than epsilon/2.
Not greater, smaller: 0 is the smallest non-negative real number!
@@Apollorion thank you. Corrected.
Great explanation.
Fun fact we in Algeria we study real analysis in the first year of bachelor degree
You can simplify the proof by using the lemma :
lim_(x->a)f(x)=L iff there exists K>0 s.t. for all epsilon >0 there exists delta s.t |x-a| |f(x)-L|
Alternative idea: Prove that the multiplication map (x,y) -> xy, is continuous R^2 -> R, which isn't that hard. Then, the result follows easily.
On second thought, it's more or less the same as your proof, but seems simpler because of the removal of f(x) and g(x).
I was just trying to remember this proof! (with some difficulty after ~15 years 😅)
This video is filled with tasty _rigor_
I love it...
Make more of these please
Omg I can’t believe I actually understood all of this!
Now proof that the proof is 1618 times harder than the calculation
..and not 1619 or 1617 times harder.
Sir if we have to prove cone curve surface area then its can be prove by like this ,we take a triangle in it and move about dy angle from central then on cover surface area there would be a thin triangle and bease would be r× dy(it approximately be taken as straight line in calculus)and height lateral height L then area of that triangle would be 1/2×r×dy×L and when we intergrate it with limits 0 to 2pie our answer is pie ×r×L, sir does this is correct , and dy×r is straight line
The example needs not the propiety, the function is continous therefore the limit converges to the value
Using 1 is nice. I used to introduce another constant something epsilon2
PF of "as x→a, f(x)/g(x)→L/M" directly is also "hard".
"directly" i mean without proving the special case "as x→a, 1/g(x)→1/M" and using prada law for f(x)/g(x)=f(x)*[1/g(x)]
Do more proofs please. ❤
awesome!!
I liked how U say ah ah fells so good😂😂😂
Woahh......I feel smarter
Zero pairs are evil period for me - this is why I liked when Dr Peyam did the proof for derivative of product rule he drew a bunch of quadrilaterals within quadrilaterals and subtracted them out. - only caveat is you need to suspend disbelief that this does not work just for positive but negative area (and functions as well).
Please make a tutorial on how you switch your markers on this board
I didn't know you were a genius in humour too mister
@@epsilia3611 thank you 😅
waiting for someone to buy me another apple pencil lol
@@blackpenredpen 😂
Hi Blackpenredpen. Make a video demonstrating why π is irrational. I am subscribed to your channel, very good videos. Greetings from Mexico
Here is how I broke up the functions:
|f(x)g(x)-LK| < ε3
Add and subtract 2LK, and add and subtract [Kf(x) + Lg(x)] to get
|f(x)g(x)-Kf(x)-Lg(x)+LK+Kf(x)-LK+Lg(x)-LK| < ε3
|[f(x)-L][g(x)-K]+K[f(x)-L]+L[g(x)-K]|< ε3
From here, you break it apart using the triangle inequality which gives eventually
|f(x)g(x)-LK|< ε1ε2 + |K|ε1 + |L|ε2
Using nonstandard analysis, this theorem becomes quite easy to prove:
Start with lim(x->a) f(x)g(x) =: A.
For any nonzero infinitesimal h, we have
A = st(f(a+h)g(a+h)).
But since
L = lim(x->a) f(x) = st(f(a+h)) and
M = lim(x->a) g(x) = st(g(a+h))
for any nonzero infinitesimal h, there are infinitesimals k,l for which
f(a+h) = L + k and
g(a+h) = M + l.
This directly implies
A = st((L+k)(M+l)) = st(LM +kM + lL + kl) = LM.
QED
I feel like its kind of the engineer way how you explain the first example. A (for me nicer way of justifying this is by setting: h(x)=f(x)g(x), and assume they are continuous, which implies, that the limit for all a (exept infinity) exists. Now we can use sequence continuity by saying: lim ( f(x)g(x) )= lim h(x)=h( lim x ) = f( lim x )g( lim x). And we are done. I am of course assuming that you need to know about sequence continuity, but this should be more elementary than limit to some random point a, where you often have some 0/0 or infinity/infinity problems :)
Thanks PROF)
26:30 I think it would be a clearer explanation to simply state that for all values of L, |L| < |L| + 1. Diving both sides by |L| + 1 gives [ |L| / (|L| +1) ] < 1, so you have [ |L|*eps / 2*(|L| + 1) ] < eps/2. Great video! Haven’t seen this proof since my undergrad in 2018.
That’s what i was thinking too. If we regroup the terms as
[eps/2][|L|/|L|+1]
Next video: 100 matrix transformation and eigenvalue questions
this is very easily taught in asia. its just that NA students arent exposed to inequality too much.
Back in the day when I used to have to do such proofs, I was irritated by the need to "edit" the proof after completing it, so choosing epsilon/3 for example, as the last line added three such terms. It makes the proof more "magical" for those trying to read it ("Why epsilon/3!!!!") After all, anyone who understands the limit/continuity/whatever definitions understands that "< 3*epsilon" is as good as "< epsilon" given the arbitrary choice of epsilon in the first place.
Lol I was lazy and sometimes said you see how all the lines of deduction are in order when thinking from scratch and stuff - told the professor in comments please read from bottom to top in reverse. They said no - you have to rewrite the entire proof again. I found that was annoying.
Can you make Differentiation and integration basics concepts for beginners.
I'm unable to understand it in my school.
Hi dear Professor, I have a topic that might interest you and I would like to see a video about it:
PROVE THAT FOR EVERY REAL NUMBER NOT A MULTIPLE OF 2 CONTAINS IN ITS DECOMPOSITION AT LEAST ONE ODD NUMBER
Once you have delta-1 and delta-2, do you even need delta-3? That is, can’t you take min(delta1, delta2) and leave it at that?
Yes, bc I used the given limits for three different ineq in the blue part.
What is the name of the whiteboard program??
I remember when I first watched bprp I was in high school and didn't know many things. Now I can be more rigor than bprp.
Could you prove that integral of e^(-x^2) has no solution in elementary functions?
try finding the values for a, b and c when
f(x)=ax^2+bx+c
f(A)=B
f(C)=D
f(E)=F
I did this equation myself and the answer is very long
The last step can be easily justified by noticing
ε|L|/[2(|L|+1)] = (ε/2)[|L|/(|L|+1)]
and also
|L|/(|L|+1) < 1
This is true even when L=0. Therefore
ε|L|/[2(|L|+1)] < ε/2
QED lol
Edit: I realized afterwards many others explained this already or similar arguments haha but yeah the proof is sound, was really nice following it till the end
Give suggestion to study PDE , Its hard to grasp its concepts
I don't remember L'Hopital's Rule being so difficult needing f(x)/h(x) when there is a numerator and denominator limit of the form f(x>a)g(x>a) 😬🤣
Blackpenredpen can you do a proof of descartes rule of signs. I'm starving for that proof. You would explain the proof very easily. 😢
Im willing to pay you
1+1=2 also hard to prove if you look in uni books
How about:
dy/d(dy/dx) = y
if |L| = 0 the expression is 0 and is definitely less than epsilon/2 (because epsilon is positive and half of it is still positive), otherwise |L|/(|L|+1) is less than 1 so the second expression is also less than epsilon/2.
I'll ask you a question
What is -ln(-1) =?
A . iπ
B . -iπ
C. A ans B both
D. Can't possible
E. None of the above
I am not testing you but by watching just your videos i got that question in my mind
When I did an online real analysis class the professor only proved this for limits of sequences. Then after that he didn’t bother proving it explicitly for limits of functions, rather he used the sequential characterization of limits to be like “we did this already” lol. I don’t blame him for not wanting to go through this again honestly!
Also for the |L|/(1+|L|) thing, you could also break it into cases, ie if |L| = 0 then of course ε|L|/2(1+|L|) = 0 < ε/2. Then in the case |L| is nonzero you do what you showed in the video.
In fact, I should have done it as (like many people who have already pointed out)
|L|/(|L|+1) < (|L|+1)/(|L|+1) = 1
I couldn't believe I didn't think of that when I was working this out lol
@@blackpenredpen honestly as a grad student in math, I feel like that a lot 😂😂😂 I’ll spend hours trying to come up with a certain proof and then when I look at my professors solution I’m like “why didn’t I think of that” hahaha
Let ɛ < 0
We allwayse proof the limits using the defenition of limits
0:45 Does the limit of x as x approaches 3 exist? Yes, it does. How do we know? Epsilon-delta time!
This is why we have these theorems in the first place, right? You don't want to go through this ε-δ stuff every single time, so you go through it *once,* to prove the theorem, and afterwards just use the theorem.
@LeftRight : That's true, but even those are closer to the general proof for multiplication that we did, than to the specific proof for x√(x+1) that we avoided. (Besides, if we really prove all the theorems that calculus students implicitly rely on, for addition, multiplication, powers, trig functions, etc, then we'll have gotten plenty of practice.)
What videos is he referring to, where he has explained the epsilon-delta.
czcams.com/video/DdtEQk_DHQs/video.html
See links in description 😃
Nice proof! We can also let |g(x)-M|
27:00 this part is simple. Notice |L|/(|L|+1) < 1
Great explanation. Technical requirements explained so well that a 14 year old could understand it 👍.
Glad to hear 😃
Sir,I try this as a undergraduate(12th) student but I can't. And also want to prove derivative of u/v and uv rule. Love from India.
👍
Please integrate this hard problem :
İntegral ln(x)ln(1-x)/(1+x²) dx borders 0 to 1 😓😓
I always though this was obvious,and therefore non provable😅
Now do the Chain Rule 😅
Can I ask, how do we use the epsilon-delta defn to prove a limit than involves x-> +/- infinity? Because infinity is not a number so we cant say a=+/-inf. Or is there a different definition we have to use?
Thank you in advance
I will give an example
Say we need to prove lim x->+inf 1/x = 0
We need to show that for all epsilon > 0
There exists delta st. For all x>delta(THE KEY DIFFRENCE)
f(x) - 0 < epsilon
One way to see this is by substituting y=1/x and take the limit y->0
@@thatapollo7773 Thank you. But it should be |f(x)-0|
Yes, there are different definitions for all of those. For example, if you want to prove lim[x->a]f(x) = +infinity, you have to show that for any M there exists some delta>0 such that if |x-a| < delta then f(x) > M. It’s similar to epsilon-N definition. It’s a bit different when a = +/- infinity and all the other versions but they’re all basically the same.
@@Ninja20704 yes, I forgot to put the absolute value
PROF thinked this better than blackboard.
There is a subtle mistake because just because the function f and g converge to certain values L and M that doesn't mean there continous only if the value of the function is the same as the limit so L should be f(a) and M=g(a) but okay. shorter proof: Because f and g are continous they're bounded in a little neighbourhood around "a". We can find a bounding constant for both of them: ∃M>0: |f|,|g|0 and because f and g are supposed as continous functions we find a delta for both of them such that |f(x)-f(a)|
tried that during class. wasted 30 minutes, not sure if my proof works, there might be an easy proof that doesn't even rely on delta epsillon:(. guess imma post my proof on reddit and hope for the best.
way better than whiteboard
Next video: Polynomial division?
"The proof is left as an exercise for the professor."
If |L| = 0 then obviously |L|*|g(x)-M| =0 since this is 0 multiplied with something. And 0 is always less than ε/2 since ε is positive. For |L| >0 the shown method works.
1618 = 2×809
😂Is that a Monty Python joke? 1618, a movie about the Spanish Inquisition?
ETA: |L|/(|L|+1) < 1 since |L| < |L|+1. No shenanigans necessary. Great video.
No, 1618 is 1000*φ, the golden ratio 😃
@@blackpenredpen Ah, nice. I didn't expect that or the Spanish Inquisition.
10:18
for me the worst in maths is when you have to prove an evidence (1+1=2)... Great video by the way
Thanks!
Why do you always indicate that |x-a| > 0? Is that only to specify that x=/= a?
Well, x ≠ a is equivalent to |x-a| > 0, which is part of the definition of a limit.
Bring a calculus intro video for a 8 grader
This actually pretty a basic proof you could even do it in a normal Vector Space it's the same thing