Don't use adblocker... it demonetizes his videos. If at possible watch the ads and let them play through and show him some support. Beside I don't mind them spending money on a product I'll never buy to support ones I would pay for.
I don't know what the difference between my textbook and my professor and you is, but something just clicked watching this. Thank you so much, you are saving my grades.
Hey. I now understand how this relates to the overall picture of limits definition that is for every epsilon we have a delta, but things would be even better if we approached this problem from a geometric perspective: getting delta from a graph. And thanks for your videos!
I feel so bad because I finished all three CALs and now I am in differential equations, but I was never able to master this proof. I just hate the stupid epsilons man.HAHAHAHA. Thank you for the video Patrick.
I regrettably feel that after finishing and currently studying differential equations, I was never able to fully understand this proof. Thank you to my advisor for pointing me in the right direction.
It seems [to me] that this epsilon-delta strategy can only affirm a limit, if the function behaviour is well behaved around the point of discontinuity. Which presupposes prior knowledge of the behaviour of the function. That would make the proof, self serving, in that it aims at proving a property thats already evident. By well behaved, I mean that the function is either sequentially increasing/decreasing on either side of the proposed limit. What about irregular functions that make jumps in F(x) as we get closer to the proposed limits? I really don't know, so I need some coaching. May be it would be helpful if we began with a definition of the types of functions where this approach can work or not. TYSM for the vid
Indeed! This is why I couldn't get it. I was like, why in the hell are we doing the same thing, but in reverse order?! of course we're going to get the same thing!
Thanks for all the videos bro. Listen, can you do an example where the inequality cannot be solved algebraically, so you have to "enlarge" the function in order to get something that CAN be solved, then from there prove that the enlarged function is less than epsilon, which proves that the original function we started with is ALSO less than epsilon since its smaller than the function we enlarged. I hope I was able to explain what I'm asking.
@@bilaltariq7819 you neglect to mention that epsilon is totally arbitrary and delta DEPENDS on epsilon....and how crucially important that is....but still FACTS. Lol for example if you have a function f(x) = 2x 2x = 10 < 11 and 2x = 4 < 11 SURELY, you ain't gonna say 10 = 4 lol.... It's more like f(epsilon) = delta so to speak. So since f(epsilon) = delta we have f(epsilon) = (1/2)E = E/2 = delta. If he didn't work the other way, it would be a prank. lol A complete s c a m.
Don't get confused by all the technical terms guys. The simple intuition behind limit is the value approached by a function as the input approaches a certain value. That's all, the epsilon and delta definition is stating the same thing. As long as you understand the meaning of what a limit is, you don't need to know the technical confusing terminology.
1st : absolute of function is less than e and prove absolute of points less than a manipulation ( related to e which is dell ) 2nd : now take that absolute of points less than dell (but put manipulated value) will always implies absolute of function is less than e
My professor did it by getting a inequality for f(x) - L from the delta and then he chosed a ephsilent for that delta is it correct? Sorry for my english
I am confused about the part where we let delta = epsilon/2. Why does this work because wouldn't substituting ABS x-4 make it so that delta is less than epsilon/2
What I learned: if 1+1=2, then subtract 1 from both sides: 1=2-1 1=1, then let 1 equal to 1, 1=1 add one to both sides: 1+1=1+1 which is also equal to 1+1=2 BOOM here you go, enjoy your accomplishments, you've proved absolutely nothing.
I don't get why this isn't a tautology. You basically did some math, then you did the opposite steps to get back to where you started and somehow that is a proof?
I know its been awhile since you posted this but the reason delta = epsilon divided by 2 is simply that we are letting delta = epsilon divided by 2. We wanted to find out what we needed to set our delta value equal to so that we could get | 2x + 3 - 11|. If you ever take Calc 3 this will still be useful to know when taking limits in multiple variables.
I understood everything except for the part where you said we could use epsilon/3 or epsilon/4 and it would still work. At that one moment you completely lost me
To prove that this limit exists and satisfy the definition, we need to establish a relationship between δ and ε such that, if x is within δ of a, f(x) is within ε of L. For example, say we find that δ is ε/2 for a particular limit, and say we arbitrarily pick 4 for ε. This means that, in order for f(x) to be no more than 4 away from L, x must be no more than 2 from a. If we change δ to ε/4, this definition is still satisfied because for an ε of 4, we have a δ of 1, and an x no more than 1 away from a still produces an f(x) no more than 4 from L. We can make δ whatever we want as long as it is not more than ε/2. Basically, we can make the tolerances for δ as strict as we like, but we cannot make them more lenient. If δ is greater than ε/2 in this example, f(x) will no longer be within the desired distance of L, and the definition cannot be satisfied.
Well, I kind of started getting it just now, and I've attended around four Calculus courses... so keep trying and someday it will make sense (that's not supposed to sound pessimistic).
So hard to see anything with your hand in the way. I had to keep rewinding to see what was under your hand. What should have been around a 7 minute video took me nearly 30 minutes to watch. Also, there's only one example and more are needed to fully understand this topic.
Want to say goodbye to the ad"what! still searching math help from youtube?"
yes, i am still doing that and it is really helpful and free
lmao that ad pissed me off so bad.
Oh and Use adblocker.
Don't use adblocker... it demonetizes his videos. If at possible watch the ads and let them play through and show him some support. Beside I don't mind them spending money on a product I'll never buy to support ones I would pay for.
Exactly
7 months later and I get the same ad
Just got it now..
Thank you so much, you are a godsent, my Calculus 1 Professor is making me learn this on my own and you just saved my life
relateable
Fr in that situation rn 😭
I feel you
I don't know what the difference between my textbook and my professor and you is, but something just clicked watching this. Thank you so much, you are saving my grades.
DUDE THANK YOU SO MUCH THIS IS ABSOLUTELY CRAZY THIS MAKES SO MUCH SENSE NOW.
Glad it helped!
Thankyou so much, my professor didn't do a great job of showing this and now it all makes sense!
Hey. I now understand how this relates to the overall picture of limits definition that is for every epsilon we have a delta, but things would be even better if we approached this problem from a geometric perspective: getting delta from a graph. And thanks for your videos!
lol it took me a month to figure out derivates means rate of change at point.
@@blacktape52black72 lol XD
I feel so bad because I finished all three CALs and now I am in differential equations, but I was never able to master this proof. I just hate the stupid epsilons man.HAHAHAHA. Thank you for the video Patrick.
I really don't know what I would do without you sir. Thank you ver much...
I regrettably feel that after finishing and currently studying differential equations, I was never able to fully understand this proof. Thank you to my advisor for pointing me in the right direction.
so clear! never understand the concept of precise limit but u just cleared that out
It seems [to me] that this epsilon-delta strategy can only affirm a limit, if the function behaviour is well behaved around the point of discontinuity.
Which presupposes prior knowledge of the behaviour of the function. That would make the proof, self serving, in that it aims at proving a property thats already evident.
By well behaved, I mean that the function is either sequentially increasing/decreasing on either side of the proposed limit.
What about irregular functions that make jumps in F(x) as we get closer to the proposed limits?
I really don't know, so I need some coaching.
May be it would be helpful if we began with a definition of the types of functions where this approach can work or not. TYSM for the vid
thank you so much. I was thinking I'm never gonna get this concept ..but after watching this video, I feel like I got it.
Thank you so much! Wayyyy better explanation than my professor.
This video saved me from poor instruction. Thank you sm
But HOW did you decide that delta=epsilon/2?
This video explains the precise definition better than the other one you showed. thx ✌️
Shouldn't it be 0 < |x-a| < δ, not just |x-a| < δ, to ensure x≠a?
Yes, that's true. He probably didn't write it to simplify things.
Thanks dude, got exam tomorrow and it really helped
If only I had this last week or my calc1 test!
Wow, perfect timing to post this...my first calc1 test in a few hours and this proof is the hardest concept of the first section.
Yo how did the test go?
@@uyujnnqaal lol I ended up dropping the class, didn't need this level of calc for my major it turned out
Saying ' just reverse the steps' is very confusing--you should focus on the implication issue. great video!
Indeed! This is why I couldn't get it. I was like, why in the hell are we doing the same thing, but in reverse order?! of course we're going to get the same thing!
Thanks for all the videos bro. Listen, can you do an example where the inequality cannot be solved algebraically, so you have to "enlarge" the function in order to get something that CAN be solved, then from there prove that the enlarged function is less than epsilon, which proves that the original function we started with is ALSO less than epsilon since its smaller than the function we enlarged. I hope I was able to explain what I'm asking.
oh my god i love u😭i was struggling so hard
I can't express my gratitude 💕💕💕💕
Thank you very much sir!! Your channel helps and motivates mi in school 😊
happy to help you :)
Hello can you reverse
|x-a|
i don't understand why E/2 = DELTA.
for example:
6
@@bilaltariq7819 you neglect to mention that epsilon is totally arbitrary and delta DEPENDS on epsilon....and how crucially important that is....but still FACTS.
Lol for example if you have a function f(x) = 2x
2x = 10 < 11
and 2x = 4 < 11
SURELY, you ain't gonna say 10 = 4 lol....
It's more like f(epsilon) = delta so to speak.
So since f(epsilon) = delta
we have f(epsilon) = (1/2)E = E/2 = delta.
If he didn't work the other way, it would be a prank. lol A complete s c a m.
Thank you very much Mr Patrick. Now am confident enough that i will muster calculus!!!
Don't get confused by all the technical terms guys. The simple intuition behind limit is the value approached by a function as the input approaches a certain value.
That's all, the epsilon and delta definition is stating the same thing.
As long as you understand the meaning of what a limit is, you don't need to know the technical confusing terminology.
How can we let delta equals to epsilon by 2 ??
I mean in what sense is it supposed to work??
Please reply...
Super explanation sir I could really understand your lecture
In what part of the limit definition it´s said that the point can´t be L? or a ?
i subbed and it was worth it. Thanks Mate!
Why epsilon over 3 over 4 and all will work....???please help
1st : absolute of function is less than e and prove absolute of points less than a manipulation ( related to e which is dell )
2nd : now take that absolute of points less than dell (but put manipulated value) will always implies absolute of function is less than e
Here what is meant by Delta and epsilon exactly .can we take any other letter instead of them as which means small +ve number
My professor did it by getting a inequality for f(x) - L from the delta and then he chosed a ephsilent for that delta is it correct? Sorry for my english
When I look through my textbook, they randomly get values for delta and then I get confused because idk how they did it
and now you know?
Did you use Differential and Integral Calculus by Love & Rainville as a reference for this video?
Watch from 3:40
I am confused about the part where we let delta = epsilon/2. Why does this work because wouldn't substituting ABS x-4 make it so that delta is less than epsilon/2
becase abs x-4 = delta.
Great video thanks man!
What if you get a problem where a is a negative number?
THIS IS GEM! THANKSSSS!
the only part im confused on is why youre allowed to say L = E / 2 ?
what about the pre-proof?
Thanks! I love your video!
NOW I UNDERSTAND THANK YOU!
YEAHH NOOWWW I UNDERESTAND IT THANK YOU LETS GOOOOOOOOOOO
plz can you show how to prove lim 1/(1-x)^2 = infinity x tends to 1
Sangat Membantu, Terima Kasih ( Indonesia) = So helpfull, Thank You
this video helps me to pull an all-nighter cause my (smart) lecturer :D
Thank you buddy❤
Patric u are amazing i've learnt things from u in two nights watching u is also enjoyfull.u deserve fucking oscar prize
Hey Patrick! Just curious, is this your main job, or do you have a day job?
Nice explanation.
makes sense that a left handed person is tutoring calculus XD
this is beautiful
what do you mean prove it just does
bro can you tell me express this definition of limit using quantifiers
helps a lot!!
thanks!
thx bro, IM FROM TUCKER.
Im awestruck
The philosophizing is necessary
Tqsm sir 😍
nice but can you please make a video to help me do better in algebra one thank you
thank uh so much sir🥰
Most welcome
@@patrickjmt can you plz make more vedios on this topic🤗. Love from india
Thank you
i want to get solution of the end
Give a smart guy a sharpie and a camera, and he will do wonders!
Thank You! :)
Thank you a lot lot lot lot lot lotttttt
I sure love that i'm taking calculus as a biology student.
But, this was actually very helpful.
You should be taking Calculus if you are a Biology student though...
Thx Man :)
Hi,What are u study?
What I learned:
if 1+1=2, then subtract 1 from both sides:
1=2-1
1=1, then let 1 equal to 1,
1=1
add one to both sides:
1+1=1+1
which is also equal to 1+1=2
BOOM here you go, enjoy your accomplishments, you've proved absolutely nothing.
Are you still there?
Well just wanted to say that you are the only comment that made me laugh in this horror movie
Thx for it
it would help if you could show a case where the limit DOES NOT exist - eg at a step discontinuity.
very nice
Thank sir
Cool thank you
Praise
thx
I don't get why this isn't a tautology. You basically did some math, then you did the opposite steps to get back to where you started and somehow that is a proof?
a?
Thank yooouuu!
Güzel anlatım 👍
Bizim yarak kafalı hocalardan bin kat iyi
@@vahsibatljohnny6682 😂
im confused, how is it mathematically proven that "delta" is equal to "epsilon dived by 2?" thanks guys
I know its been awhile since you posted this but the reason delta = epsilon divided by 2 is simply that we are letting delta = epsilon divided by 2. We wanted to find out what we needed to set our delta value equal to so that we could get | 2x + 3 - 11|. If you ever take Calc 3 this will still be useful to know when taking limits in multiple variables.
It just makes it easier to solve
god bless
AMAZING
😄😄😄😄😄😄😄
quentin tarantino?
DUUUUUUUUUDEEEEEEEEEEEEEE
You are reading from thomas calculas
I understood everything except for the part where you said we could use epsilon/3 or epsilon/4 and it would still work. At that one moment you completely lost me
To prove that this limit exists and satisfy the definition, we need to establish a relationship between δ and ε such that, if x is within δ of a, f(x) is within ε of L. For example, say we find that δ is ε/2 for a particular limit, and say we arbitrarily pick 4 for ε. This means that, in order for f(x) to be no more than 4 away from L, x must be no more than 2 from a. If we change δ to ε/4, this definition is still satisfied because for an ε of 4, we have a δ of 1, and an x no more than 1 away from a still produces an f(x) no more than 4 from L. We can make δ whatever we want as long as it is not more than ε/2. Basically, we can make the tolerances for δ as strict as we like, but we cannot make them more lenient. If δ is greater than ε/2 in this example, f(x) will no longer be within the desired distance of L, and the definition cannot be satisfied.
delta = 0.00000000000000001 and epsilon = 0.000000000000000002 make a tiny difference.
thank you for the explanation. Please use a normal pen or pencil while writing on a paper
can't even watch this with his hand in the way of everything written had to pause like 125 times ... leftys
This epic
god this is never going to make any sense to me. i give up on precise definition of a limit. this is just totally beyond my INTELLIGENCE.
Well, I kind of started getting it just now, and I've attended around four Calculus courses... so keep trying and someday it will make sense (that's not supposed to sound pessimistic).
😸
So hard to see anything with your hand in the way. I had to keep rewinding to see what was under your hand. What should have been around a 7 minute video took me nearly 30 minutes to watch. Also, there's only one example and more are needed to fully understand this topic.
wtf
Hindi me bolo hard eng heijauchi
Markers on paper is making my ears bleed
oh, the horrors