Formal definition of limits Part 4: using the definition | AP Calculus AB | Khan Academy
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- čas přidán 10. 01. 2013
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"let me switch colors just to ease the monotony"
Whenever a point x is within δ units of c, f(x) is within ε units of L
This actually helps a lot, thank you.
Not going To lie this is exactly what I wanted to know
Thanks Sal my prof moves so quick I don't understand him at all. You helped me figure out the whole concept.
If you guys don't get it, watch the other two videos he has for background info. They help it all make sense.
Sal Khan absolutely love you and the work you do. You helped me from being a high school math failing kid, to someone now majoring in math.
God bless your soul! Jesus, you have no idea how thankful I am. This shit was so confusing when my prof explained it but you made it so easy to understand!
What are you doing right now?
@@berkebayrak9198 what are you doing right now?
As a first-year engineering student who just understood what might be a problem on my next week´s calculus exam, thank you
Same here. I just had my first class today and didn't get a word my prof was saying i just wrote the notes down but I kinda see it now. good luck on your exam!!
@@jjm8677 me too. how are you doing rn?
This video is mind blowing..I don't have words to express what I felt while watching this video. You're so good at explaining things.
I would fail my calculus class without this channel
An entire chapter summarized in 8 minutes. Thanks.
delta epsilon hurts :(
I was good at math until epsilon delta. Just can't get it and this seems to not help me at all.
Same here bro, don't worry,but now I am just beginning to understand a little better.
Todd Collins Think about it like this: The letters ε and δ can be understood as "error" and "distance". ε as an abbreviation for error. In these terms, the error (ε) in the measurement of the value at the limit can be made as small as desired by reducing the distance (δ) to the limit point.
+Todd Collins Was good until uni math :'( day 23, still haven't seen a number yet. Help I'm dying here :(
just think there's a distance between y value and limit that maps to between x and a values
When i get into something that i dont understand in math, i ussualy try to answer questions like "Well, what does this part of the equation tell me?" i think that it's of no use if you know the epsilon delta definition , or any other thinkg, if you don't understand what is expressing, it's like translating between two languages, you have to understand the meaning of every word and the rules of the language to translate its meaning
Hey Sal, this is really a lifesaver; I had no idea what this concept was until I saw this video, and now I know exactly what was taught to us for this. BTW, we have this definition only for multivariate calculus, not for single variable calc, for some reason.
Your videos are very precious to me. Had many problems in my life during high school so I struggled with maths but the information you provide online helped me to fill my gaps and I am grateful
Great! Powerful to learn the fundamentals of proving limits with the epsilon-delta definition.
I dont have words to thank you. I was working so hard to learn this but didnt understand anything. Thank you so much for this.
three cheers for Khan Academy!
Wow really good explanation ❤!
I have watched every video in this playlist.. EVERY SINGLE ONE OF THE 54 VIDEOS!
Achievement unlocked: ATHEIST
I love this man.
I've finally understood. :)
YASS MATH GOD THANK YOUU
It's 4, because 2 is in the domain of your function and (therefore) you can simply plug in 2 for x in your original function (x^2):
2^2 = 4
Therefore, the answer is 4.
You are right and implicitly implying continuity to prove this limit. Now if you want to show this without continuity and using epsilon/delta definition, how would you do that? a video from Sal would be good!!
Wow!!! Nice explanation.
thank you sir it helps a lot
Thank you that was helpful
very helpful video, thank you
Outstanding video lecture.
thanks man!
Very good explanation, thank you sir.
You guys probably should learn calculus from Spivak's book, if you haven't already. For anyone serious with math :)
Well, now I have a headache but I think I get it...
How can I say (in this example), that δ = ε/2? I understood the inequalities, but....how can this equality be true? What makes me sure that this equality should be?
Epsilon is arbitrary, so you can let it be any positive real number.
khan u r extra ordinary
very useful
now i see whre im goin thanx s'much
what do you do if the limit approaches infinity? I am given the problem to find delta for the limit of (1/(8-x)) as x approaches 8 from the left = infinity.
I CANT BELIEVE THIS HELPS PEOPLE
You are the best
Doesnt this simply mean that for a change in x , there will be a change in y by some value which should obviously be gradient, according to which in this case, where delta = absilon/ 2, for any change in x around 5, there will be change in y by 1/2 which is = gradient
Isnt this just a complicated way of defining gradient???
This may help; obviously there's nothing intuitive here, but just think of ε as being incredibly small and 1/ε as, consequently, being incredibly large. We say that the limit of f(x) as x→w does not exist (or equals complex infinity) if & only if for any ε > 0, there is a δ > 0, such that for all x in the domain of f(x) with |x - w|
I am in Calc 3. i have attempted to do epsilon delta problems since calc 1. i still dont get it
Thanks
This was an unorthodox solution.
| f(x) - L | < epsilon becomes | 2x - 10 | < epsilon. 2 | x-5 | < epsilon.
| x - c | < delta becomes | x - 5 | < delta.
So, we simplify the epsilon equation by dividing by 2.
| x - 5| < epsilon/2. therefore delta
lol. idk if this was a joke, but it means "Quite Easily Done"
It's Quod Erat Demonstratum if I recall correctly, is some Latin meaning "I have proved what I set out to prove." Kinda a classic after a proof.
i have a problem similar to this however i dont have the equation, just the graph with tje x valie and epsilon. i need to find the two delta values, how?
Did you watch the other 2 videos? They help explain it further...
still dont really understand why it needs to be represented in such a complex manor, isn't all this just like saying if F(x)= y then y/F=x so if you want the delta (distance from x) of a respective epsilon (distance from y) then you just do (epsilon)/F= delta.
or with a non liniar just plug in the respective y values which corespond to y- or + epsilon to get the respective x + or - delta
but what when x approaches infinity ?
I've been taught that you only use this definition to prove the existence of a limit. To prove that a limit does not exist you should use the heine definition of a limit.
Great
What about limits at infinity?
What I don't understand is why there has to be an epsilon or delta in the first place. When taking a limit in 2D I don't remember being concerned with any sort of domain around the limit L. Why can't delta and epsilon be out to infinity if all we care about is what happens close to the point P(x,y)?
+MisterBinx How would you prove the existence of the limit in that case?
You should do an example where the limit does not exist for a certain x value.
Why can't x be equal to c? Given that -δ < 0 < δ will hold true.
I agree
OMG HEY DEE!!!
I followed until "|x-5| < delta" at ~3:40. The video seems to simply state this as fact without explaining how this is arrived at.
I found this is actually explained in the previous video. "|x-5| < delta" is related to "|x-c| < delta". The way "|x-c| < delta" is arrived at is writing the phrase"x is within delta of c" as an inequality.
...since when is it possible that
lim x->5 f(x) =/= f(5)?
Concretely, what avoids the supposed contradiction?
Is it still correct if I apply the exact same proof to a non-linear function ? Like x^2 for exemple ? If I try to do it, I obtain delta as a function of epsilon AND x. I don't see any problem with it, but I would want to be sure about that.
Thanks
+chocochoco232 Just x^2 for all values of x? So totally continuous? Because in that case, the (epsilon,delta)- definition works slightly different. You get the expression |f(x) - f(c)| < epsilon, since f is defined in c itself. Tell me what you were thinking about with just x^2.
Wouldn't that be 4 as when you evaluate the function f(x)=x^2 at 2 it equals 4. Maybe if the function was f(x)= x^2, x=/=2 and 2, x=2? Please ignore the second part if it was no help.
a good example to try is f(x)=(1/x)
What if function f(x) have a horizontal asymptote y=k? In that case, for every epsilon which is larger than |k - L|, what will the corresponding delta be? Or the limit for f(x) as x approach to c just simply doesn’t exist? For example, f(x) = 2^x/ (x+1) can you claim that [ lim x -> -1 f(x) ] = 1/2?
No.
So what would be the final answer?
I'm studying for my exam :_)
Easy for polynomial ftns....becoming difficult for complex functions.
epsilon mega delta less than delta epsilon two which is equal to epsilon less than delta/2 epsilon epsilon delta over epsilon equal delta over delta by epsilon which is epsilon???
where is part 3?
please , Sal' apart , if you (actual reader) know something about it , I'll really appreciate your help
im in 8th grade and i am doing this for school
still dont understand...
+Aulia Muhardi Arifin Try this: Work the problem on paper while he is explaining it, pausing the video after each step. If the step or point that he covers doesn't make sense, watch that part again until it does. Also use this method with other functions, referring to this video when needed to find connections. I'd say it has taken me about 6 hours to really grasp this concept. Don't give up.
+Aulia Muhardi Arifin It takes time. Just keep watching video after video and reading the textbook. The pieces start falling in to place bit by bit.
Is't this just a definition of the limit. It is usefull for proving that a limit exist. There are other ways but I doubt that Sal will teach them they are said to be much harder to understand.
Sal, can you do lim x^2 as x approaches 2?
It's just 4 my guy
How to define limit such that x is approaching infinity and f(x) is also approaching infinity using epsilon-delta ? HOW ?
Simple, you dont, because infinity has no limits.
YourGamingPill why ??
how can you define a limit for something that has no end? infinity is an endless figure.
urgh, it's not about defining what is limit . hmmmm.
en.wikipedia.org/wiki/Limit_of_a_function#Infinite_limits
so, is the limit correct? as x approaches 5 will f(x) be equal to 10?
Just found out Khan was VladTV
Wait, what exactly does the epsilon represent?
From what I understand, it's the difference between a random number next to the function's limit and the limit itself; no matter how small, it'll always be positive. Delta is relative to it.
You can associate epsilon with the Y axis of the graph. We're talking about the possible Y values between f(x) and Epsilon. If i'm not mistaken ahhaahah
Yeah epsilon is basically the difference between f(x) and the proposed limit. i.e.
| f(x) - limit |
So when we say | f(x) - limit | < epsilon for any epsilon we mean we can take epsilon (the difference between function and proposed limit) to be as small as we want.
If it can be as small as we want (no matter how small you pick epsilon, aka the distance between limit and function, I can pick a delta to satisfy it).
And remember as epsilon is to do with the y axis, delta is to do with the x axis. Delta is just the extra distance between what x is tending towards and what we pick x to be.
So if I can give you a delta for any epsilon you give me, the limit must be what we thought. So if you picked epsilon = 0.0000001 (i.e. the distance between the function and limit is 0.0000001), then I could give you a delta. If I can do that for all epsilon > 0 then the proposed limit must be true.
SuperYtc1 Close, but I dont think it's the difference. Epsilon is the given number, some units away from L. So the difference would be Epsilon + L and L-Epsilon. IE if your limit is 10 and your given Epsilon is 1, you will be able to find a number delta that defines f(x) between 9 and 11, but not 10 itself.
Rufi83 Epsilon is the distance between f(x) and L.
Epsilon-deltas are only used to prove that the limit does exist: not the converse.If you wanted to prove the limit does not exist simply use the different path test. If you get two limit values that are different, then the limit DNE.
Wow, a video where the guy explaining is not an Indian with horrendous accent that is impossible to understand. So rare I had to pinch myself to check it was not a dream.
Awesome video.
funny cus he's still indian
Is he Indian? Ya he is just wow
n W
he is Indian-American. His accent must be perfect as he was born and raised in the US...
hahaha hi sierra!
What happens when your limit of f(x) doesnt exist
You have to prove the negation of the original statement
the direction of the proof is wrong. he was saying for any epsilon you can find a delta.... but he does it in the reverse. the proof should start in the definition of a limit , from lf (x) - L l is less than epsilon, then from there, "translate" that expression into lx-cl is less then delta.
in other words, he should not have multiplied lx-5l < delta by 2 but rather DIVIDE lf(x) - Ll < epsilon by 2!!!!. Same result wrong procedure of proof you have done here.
I agree. He should'nt assume the conclusion is correct first, rather he should've started with the assumption that IF the function for some x is a distance epsilon away from the limit value, THEN there is a delta such that any x in the interval x +- delta can be plugged into the function and be at maximum within an epsilon distance away from the limit.
кто мне все загружает я не нуждаюсь тем более на китайском шышыга отвали
i have never been so confused as someone speaks English, that is not English, its mathematics in Spanish or French or something
horrible explanation. i don't get how to get |x-5|