This was my absolute worst nightmare while taking Modern Analysis in college! But many thanks for walking us thru the steps of how this proof came to be.
As an engineer working on engineering precision mathematics we find the x equivalent in like time measurements all the time when 2x + 1 approximates to 2x with no significant error in precision. 3x + 4 would need a larger x to equal the same low error approximation as 2x + 1 approximate 2x replaced as 3x + 4 approximately is 3x for x >> 1 for large x. For example, x = 10^4 then the function approximates to 2x/(3x) or x in numerator and x in denominator show we are so close to 2/3 no matter if you have a ripple 1 in the numerator and a ripple 4 in the denominator compared to the 2x value / 3x value ... Not the 2 value / 3 value limit mistake in thinking.
For epsilon-delta proofs, when speaking through the proof out-loud, it becomes way more obvious for the student if the speaker says not just “for all epsilon greater than zero, …”, but “for all epsilon, no matter how small you decide to choose it, …” - i.e. emphasizing that our intent is to “make” epsilon “smaller and smaller”. Also “given arbitrary epsilon greater than zero” -> “given arbitrary small epsilon greater than zero”.
This is an important point that often gets lost in these discussions- you can make epsilon as small as you like and will always be able to find a delta or N that satisfies the delta/N inequality, and vice versa, if the limit is in fact L. If the L is not the limit, then you will get a contradiction.
My teacher gave us some simple limit proof qns in my proofs class when teaching us proof by construction. And this was one of them. Quick question, if we wanted to prove a limit as x->-inf, do we just change to N
I think there are many ways to define it for example: lim x-> -inftx f(x) := lim x-> infty f(-x) if it exists Or using sequences: For all sequences (xn)n with xn -> -infty, then we say lim x->-infty f(x) exists if and only if lim n->infty f(x_n) exists and is the same for all sequences I think yours is probably aquivalent
Lovely video, though I would like to point out a correction that for epsilon-N it should be stated as a defined sequence lets say a_n, so rather a set of a sequence than as you said a function:) But both work fine I guess! Ty for the videos!
I have seen an example where the δ chosen was greater than ε. I was wondering would it not throw δ outside the ε window. Can the chosen δ ever be greater than ε? Thank u
I asked myself what the limit of (1+1/x)^(1/x) when x goes to infinity. I think it goes to 1 but I don't have a way to show it and when I asked Wolfram Aloha it says 1 but the Step-by-step solution kinda goes like e^(0/infinity) equals 1, which it say is the solution. Can someone help me?
Write the expression as exp((1/x)*ln(1+1/x)) As x tends to infinity, 1/x tends to 0 and ln(1+1/x) also tends to 0. Therefore, the limit is equal to exp(0*0) = e^0 = 1 You can then write the proof it in the epsilon N form.
A simpler way to calculate the limit is that 1/x goes to 0 as x goes to infinity. So the base approaches 1 and the power approaches 0. This is not an indeterminate form, so we can legitamately conclude that the limit is 1^0=1
@@spiderjerusalem4009 of course, but I’m saying if such a small N works that it’s negative should still be fine if the inequalities hold, same way N can be a random small positive real number that isn’t close to infinity
The first part of the statement we are trying to prove is "for all epsilon>0 there exists a N>0". Therefore, N has to be greater than 0 for our proof to be legit.
i^i^i^…..=x x=i^x x=e^(ln(i^x)) x=e^(xln(i)) note: (ln(i)= (pi*i)/2 for the complex logarithm principal branch, this can be observed also through rulers identity e^(pi*i) = -1 as putting both sides to the power of 1/2 results in e^((pi*i)/2)=i.) x=e^(x*i*pi/2) x/(e^(x*i*pi/2)) = 1 x(e^(-x*i*pi/2))=1 x*(-i*pi/2)*(e^(-x*i*pi/2))=(-i*pi/2) W(x*(-i*pi/2)*(e^(-x*i*pi/2)))=W(-i*pi/2) -x*i*pi/2=W(-i*pi/2) x=(2i/pi)*W(-i*pi/2) Sorry for bad formatting, am commenting on phone.
Please solve this question for me: Let f(x)=(x²+x+1)/(x²-x+1), then the largest value of f(x) for all x belongs to [-1,3]. Can you please teach me how to apply x's bound over such f(x)?
If the first thing you comment it L’Hospitals, we already know that you’re not going to get out of this video what you need to get out of this video. It’s very telling. Come back after a discrete math class then look at this video lol
My cat, and I, have watched a lot of BR's proof vids. My cat is much better than me at giving all the proofs... just copies BR's proof. I don't think my cat understands math... do you?
@@SimsHacksI think you are right on it coming before derivatives, but calculation techniques do work in proofs. Calculation techniques in proofs is just called algebra
The blackPenredPen guy is putting on weight: ... he must be eating alot of general Tso's chicken... lol.. it's alright, I eat general Tso's chicken also... 😎
Check out the εδ definition ultimate introduction 👉 th-cam.com/video/DdtEQk_DHQs/w-d-xo.html
Right 👍👍
This was my absolute worst nightmare while taking Modern Analysis in college! But many thanks for walking us thru the steps of how this proof came to be.
Literally skipped one lecture and came back to seeing this in my tutorial. Absolute Legend.
Thank you for making this demonstration waaay easier than what I learnt in school. Time to flex it on my teachers 😜
This is just what I studied for an upcoming exam, great to refresh my memory. Thanks 🙏
As an engineer working on engineering precision mathematics we find the x equivalent in like time measurements all the time when 2x + 1 approximates to 2x with no significant error in precision. 3x + 4 would need a larger x to equal the same low error approximation as 2x + 1 approximate 2x replaced as 3x + 4 approximately is 3x for x >> 1 for large x. For example, x = 10^4 then the function approximates to 2x/(3x) or x in numerator and x in denominator show we are so close to 2/3 no matter if you have a ripple 1 in the numerator and a ripple 4 in the denominator compared to the 2x value / 3x value ... Not the 2 value / 3 value limit mistake in thinking.
For epsilon-delta proofs, when speaking through the proof out-loud, it becomes way more obvious for the student if the speaker says not just “for all epsilon greater than zero, …”, but “for all epsilon, no matter how small you decide to choose it, …” - i.e. emphasizing that our intent is to “make” epsilon “smaller and smaller”.
Also “given arbitrary epsilon greater than zero” -> “given arbitrary small epsilon greater than zero”.
This is an important point that often gets lost in these discussions- you can make epsilon as small as you like and will always be able to find a delta or N that satisfies the delta/N inequality, and vice versa, if the limit is in fact L. If the L is not the limit, then you will get a contradiction.
@@yanceyward3689to be exact, for all n > N that has to satisfy, otherwise it's a limit point
Good that you gave this video I'm about to start limits in my calc course
First time I understand. Great explanation. Thanks
Man, you bring out the inner math genius in me, and I'm 62. I follow your logic perfectly
you just explained this so well, ive watched so many videos but yours made me clearly understand this. Thankyou so much!!!
Damn, I appreciate your videos even more given my professor couldn't explain it properly in 3 hours.
My uni life's first lesson was calc1 aka math1 and our proof directly started limits with epsilon delta
Welcome back man!
Could you do a video on how to do the proof backwards/both ways?
Thank you so much these videos save my life. But do you have a video on if there's an M to find instead of an N?
@@noonstar3435 I got everything you need here: th-cam.com/video/AfrnYS5S8VE/w-d-xo.htmlsi=urptcYJbQtVVkAlS
My teacher gave us some simple limit proof qns in my proofs class when teaching us proof by construction. And this was one of them.
Quick question, if we wanted to prove a limit as x->-inf, do we just change to N
I think there are many ways to define it for example:
lim x-> -inftx f(x) := lim x-> infty f(-x)
if it exists
Or using sequences:
For all sequences (xn)n with xn -> -infty, then we say lim x->-infty f(x) exists if and only if lim n->infty f(x_n) exists and is the same for all sequences
I think yours is probably aquivalent
@Ninja20704
Yes. A usual definition of the existence of a finite limit L of a real function f at -∞ is :
For all Ɛ>0, it exists N
To infinity and beyond 🚀🥳🤸♥️💫
Very Very Thank you sir ❤
Thank you!
Lovely video, though I would like to point out a correction that for epsilon-N it should be stated as a defined sequence lets say a_n, so rather a set of a sequence than as you said a function:) But both work fine I guess! Ty for the videos!
respect from bmstu
L'hospital: it that even a question?
it doesn’t actually prove the limit rigorously though
Nice video
I have seen an example where the δ chosen was greater than ε. I was wondering would it not throw δ outside the ε window.
Can the chosen δ ever be greater than ε?
Thank u
is it necessary to speify the domain of N? like N is s subset of Real numbers when we complete the proof?
beautiful
Hi, what do you do if you have a minus sign in the denominator, so you can't get rid of the absolute value??
I asked myself what the limit of (1+1/x)^(1/x) when x goes to infinity.
I think it goes to 1 but I don't have a way to show it and when I asked Wolfram Aloha it says 1 but the Step-by-step solution kinda goes like e^(0/infinity) equals 1, which it say is the solution.
Can someone help me?
Write the expression as exp((1/x)*ln(1+1/x))
As x tends to infinity, 1/x tends to 0 and ln(1+1/x) also tends to 0.
Therefore, the limit is equal to exp(0*0) = e^0 = 1
You can then write the proof it in the epsilon N form.
@@asifthatwouldeverhappen thanks
A simpler way to calculate the limit is that 1/x goes to 0 as x goes to infinity. So the base approaches 1 and the power approaches 0. This is not an indeterminate form, so we can legitamately conclude that the limit is 1^0=1
Do you have to use the max?
What’s wrong with N being negative?
"approaches infinity"...
@@spiderjerusalem4009 of course, but I’m saying if such a small N works that it’s negative should still be fine if the inequalities hold, same way N can be a random small positive real number that isn’t close to infinity
Did you get the answer to your question? Cause I have the same doubt
@@stlegendff7390 nope unfortunately not
The first part of the statement we are trying to prove is "for all epsilon>0 there exists a N>0". Therefore, N has to be greater than 0 for our proof to be legit.
When I will have access to America?
How to do it with negative infinity
Doesn't, 5/9N = e(epsilon) can also be solution?
hey blackpenredpen, can you solve
x = i^x
as in, an infinite power tower of i's.
i^i^i^…..=x
x=i^x
x=e^(ln(i^x))
x=e^(xln(i))
note: (ln(i)= (pi*i)/2 for the complex logarithm principal branch, this can be observed also through rulers identity e^(pi*i) = -1 as putting both sides to the power of 1/2 results in e^((pi*i)/2)=i.)
x=e^(x*i*pi/2)
x/(e^(x*i*pi/2)) = 1
x(e^(-x*i*pi/2))=1
x*(-i*pi/2)*(e^(-x*i*pi/2))=(-i*pi/2)
W(x*(-i*pi/2)*(e^(-x*i*pi/2)))=W(-i*pi/2)
-x*i*pi/2=W(-i*pi/2)
x=(2i/pi)*W(-i*pi/2)
Sorry for bad formatting, am commenting on phone.
Please solve this question for me:
Let f(x)=(x²+x+1)/(x²-x+1), then the largest value of f(x) for all x belongs to [-1,3].
Can you please teach me how to apply x's bound over such f(x)?
is it 3 ?
Don't we need to write the conclusion at the end?
3:36
dalil d La Hospital
Undefined : Proof Theory Graph Trayektory
I just multiply the top and bottom by (1/x).
If the first thing you comment it L’Hospitals, we already know that you’re not going to get out of this video what you need to get out of this video.
It’s very telling.
Come back after a discrete math class then look at this video lol
Can we not do taking 1/x tend to 0 if x tends to infinity
Not calculate but prove it is 2/3. So the limit definition is applicable.
You can use this definition to prove that.
Given any epsilon>0
Choose N = 1/epsilon
Suppose x>N.
Check |1/x-0|=1/x (x>0 so 1/x>0)
@@Ninja20704 thanks brother
My cat, and I, have watched a lot of BR's proof vids. My cat is much better than me at giving all the proofs... just copies BR's proof. I don't think my cat understands math... do you?
still appealing to the version in English
This chanels name starts with my favourite k pop bands name.
Lhlpital 3secs
Are you not allowed to use L’Hopital for a prove?
Of course not. That's not a proof, that's a calculation technique. Furthemore, these easy limits are usually covered before derivatives.
@@SimsHacksI think you are right on it coming before derivatives, but calculation techniques do work in proofs. Calculation techniques in proofs is just called algebra
@@photophone5574 it's allowed if you had previously proved l'Hospital's rule. Which I don't think is the case.
No you definably can’t due to the lack of proof and evidence
please stop bringing this overly used method up, notably whose validity is beyond your question. Utter tiresome.🙄
Use L'Hopital's rule what's the problem bruh😅
But this is about how to write a formal rigorous proof, not obtain the final answer, L'Hopital's rule is more like a calculation technique
Abhi aap chote ho
lhopital only helps you in the calculating, but it itself doesn’t prove anything
Bro tell that to my professor pls
0:16 😂
The blackPenredPen guy is putting on weight: ... he must be eating alot of general Tso's chicken... lol.. it's alright, I eat general Tso's chicken also... 😎
hayatımda bu kadar rezalet bir anlatıma şahit olmadıım.