I like the fact that it makes so much sense when you are also thinking on the derivative as a linear aproximation, you get the aproximation precisely in the same way, and basically when there is only one posible outcome for that aproximation, you have differetiability.
6:38 Actually you are using differentiable at this point right now, because limit of product is equal to product of limit if both limit exists. For example lim sinx/x cannot be decomposed to lim sinx lim 1/x, just to make it rigorous.
aeoexe I should have also said "because we know both limits exit" right there too. But I did mention f'(a) is a "nice number" since we had differentiability
Thank you so much! Your explanation was so clear and concise, and your enthusiasm made it so enjoyable to watch! Thanks for being my quarantine lecturer!
There exist an alternative definition of differentiable function: Let f:[a,b]→R, then f is differentiable in Xo if exist a function g:[a,b]→R continuous in Xo so that: f(x) = f(Xo) + (x-Xo) g(x), and g(Xo)= f'(xo). By that definition f is continuous in Xo because it's a sum of continuous functions: f(Xo) and Xo are constant numbers, x is continuous in every x and g(x) is continuous by definition. #YAY
The limit of a sum is not always the sum of the limit right? For example lim n->inf of ln(n)-harmonic sum till (1/n) converges. If you split up the sum into 2 limits, you get infinity-infinity, which is undefined. I think that you need to place the restriction that f(a) is defined. P.S. I know that if f(a) is not defined then the function is definitely not continuous at a, but still, you cannot split limits in general. Peace
The way I'd prove this is by saying: "If you take the derivative of a function in a point *a*, you will have the slope the line tangent to the function in that point, but you can only have the slope of line tangent the function at that point if, and only if, the point *a* exists in the function (if it is continuous at a), since you can't have the slope of nothing" For example we have the famous function sin(x)/x, we can clearly see it's not conitnuous at 0, so that point doesn't exist in the function. It's derivative is (xcos(x)-sin(x))/x^2 , and you can see that the slope of the tangent at 0 is 0/0. sin(x)/x it's not differentiable at 0 because the point 0 doesn't exist in that function! :3 Is my proof right? QwQ
I suddenly find myself very curious about how to find the area of a fractal, as in the integral of a continuous function that is not differentiable anywhere.
If I'm not mistaken, the exact area of something that is much more complicated like the mandelbrot set is an open problem. I think the area of something simpler like a kock snowflake can be written as an infinite series quite simply, then some work can be done to find the value.
The limit of a sum, is the sum of the limits. The same is true for the limit of products. My understanding is that this is true only if the functions inside the limits are continuous. (No?) But, we are trying to prove continuity in the first place. Can we prove continuity by assuming continuity first? It is clear that I am missing something important. Can you please clarify. I need this little clarification because I haven't studied calculus for almost thirty years and I don't remember much of it. (Love your channel, BTW. It brings back much of the love for math I had as a young person. So, thanks and keep up the good work.)
Alberto DeLaRaza you are close. The limit of a sum equals the sum of the limits, on the condition that the inside limits merely exist. Remember you can have the limit existing without being continuous. Also, as another comment pointed out, “the limit existing” means it must be an actual number and not infinity, which is why you can’t say lim sin x / x = lim sin x * lim 1 / x (as x -> 0).
Awesome proof, it finally made sense to me :) I know abs(x) at x=0 is the counter-example but can you also show in general that "if continuous then differentable" is a false statement or it doesn't lead to any conclusions?
man this shows why i never got proofs. whenever i read them i tend to forget what i wanted to show in the first place or what the proof is supposed to show^^
Can you help me with the following integral? Integration of x^2 * √(1+4x^2) dx It seems to me you can use integration by parts or something like trig substitution. However i get different results when doing so
@@christinabae2625 no problem. this comment is so old i don't even remember what i commented. But BPRP and I have some (friendly) jabs at each other from time to time. Ok, to be honest it is me that does all the jabbing, but whatever
Here's a nice little set of functions. f1(x) = sin(1/x). f2(x) = x*sin(1/x). f3(x) = x^2*sin(1/x). Assume for each that the function is actually piece-wise defined where f(0) = 0 in order to "plug" the hole at 0. f1(x) is not continuous at x=0 because the function oscillates between 1 and -1. f2(x) is continuous but not differentiable at x=0 because the tangent line slopes also oscillate. f3(x) is differentiable at x=0 and every other point, however the derivative is not continuous, i.e. it is not second-differentiable.
Not by definition, the definition is that when you get closer to a certain x-value , the y-value you are approaching is the same as the value at the specific point of x. If that's true for all x then it's continuous. A differentiable function is a function for which all x-values the derivative is defined
Sir, I don't get the proof..you just proved lim x → a f(x)=f(a)....where is the differentiability here?.. we are asked to prove if a f is differentiabie it is continous
i love this channel! Who says summer has to be enjoyed away from school? Not me! Math is the best!
Nole Cuber what an awesome comment! Thank you!!
I'm taking Calculus 4 and Diff EQ this summer. I can't get enough math! hahaha
"That nerd"
@@asriel522 this is indeed very correct
You might like to know about the Weierstrass Function, a function which is continuous _everywhere_ but differentiable _nowhere._
armin yes! I studied that back in my grad school life. : )
He said that the converse WASNT true
@@gregoriousmaths266 and he gave a real nice example
I like the fact that it makes so much sense when you are also thinking on the derivative as a linear aproximation, you get the aproximation precisely in the same way, and basically when there is only one posible outcome for that aproximation, you have differetiability.
I'm loving the high energy thumbnails!
6:38 Actually you are using differentiable at this point right now, because limit of product is equal to product of limit if both limit exists.
For example lim sinx/x cannot be decomposed to lim sinx lim 1/x, just to make it rigorous.
aeoexe Since it's given that f is differentiable at a, it's totally fine to split the limit because lim (f(x)-f(a))/(x-a) exists
That is what i am talking about, because he said limit of product is equal to product of limit at 6:38
aeoexe I should have also said "because we know both limits exit" right there too. But I did mention f'(a) is a "nice number" since we had differentiability
Thank you so much!
Your explanation was so clear and concise, and your enthusiasm made it so enjoyable to watch! Thanks for being my quarantine lecturer!
There exist an alternative definition of differentiable function: Let f:[a,b]→R, then f is differentiable in Xo if exist a function g:[a,b]→R continuous in Xo so that: f(x) = f(Xo) + (x-Xo) g(x), and g(Xo)= f'(xo). By that definition f is continuous in Xo because it's a sum of continuous functions: f(Xo) and Xo are constant numbers, x is continuous in every x and g(x) is continuous by definition.
#YAY
i really like your method when you teach keep going
Beautifully done. You should make a college calculus course as you are excellent at going explaining your steps
Thank you, best video on YT on this proof!
Yeah...
The limit of a sum is not always the sum of the limit right? For example lim n->inf of ln(n)-harmonic sum till (1/n) converges. If you split up the sum into 2 limits, you get infinity-infinity, which is undefined. I think that you need to place the restriction that f(a) is defined.
P.S. I know that if f(a) is not defined then the function is definitely not continuous at a, but still, you cannot split limits in general.
Peace
The way I'd prove this is by saying:
"If you take the derivative of a function in a point *a*, you will have the slope the line tangent to the function in that point, but you can only have the slope of line tangent the function at that point if, and only if, the point *a* exists in the function (if it is continuous at a), since you can't have the slope of nothing"
For example we have the famous function sin(x)/x, we can clearly see it's not conitnuous at 0, so that point doesn't exist in the function. It's derivative is (xcos(x)-sin(x))/x^2 , and you can see that the slope of the tangent at 0 is 0/0. sin(x)/x it's not differentiable at 0 because the point 0 doesn't exist in that function! :3
Is my proof right? QwQ
sapato :3 Although that is a nice graphical way to think about it, I prefer the more rigorous (and conceptually simpler) proof presented in the video.
I suddenly find myself very curious about how to find the area of a fractal, as in the integral of a continuous function that is not differentiable anywhere.
For what I read, Daniell integral should do. With that said, I do not claim to understand it.
Alfonso J. Ramos yeah, I know it's possible, I just don't understand how to find it.
If I'm not mistaken, the exact area of something that is much more complicated like the mandelbrot set is an open problem. I think the area of something simpler like a kock snowflake can be written as an infinite series quite simply, then some work can be done to find the value.
This proof was taught to me in school. So I won't watch till the end, but I still liked and commented :) beautiful proof
The limit of a sum, is the sum of the limits. The same is true for the limit of products. My understanding is that this is true only if the functions inside the limits are continuous. (No?) But, we are trying to prove continuity in the first place. Can we prove continuity by assuming continuity first? It is clear that I am missing something important. Can you please clarify.
I need this little clarification because I haven't studied calculus for almost thirty years and I don't remember much of it. (Love your channel, BTW. It brings back much of the love for math I had as a young person. So, thanks and keep up the good work.)
Alberto DeLaRaza you are close. The limit of a sum equals the sum of the limits, on the condition that the inside limits merely exist. Remember you can have the limit existing without being continuous. Also, as another comment pointed out, “the limit existing” means it must be an actual number and not infinity, which is why you can’t say lim sin x / x = lim sin x * lim 1 / x (as x -> 0).
Ah, I see my error now! Thank you for the clarification!
"The limit of a sum equals the sum of the limits": shouldn't we all so add the condition "for a sum consisting of a finite number of terms"?
Awesome proof, it finally made sense to me :) I know abs(x) at x=0 is the counter-example but can you also show in general that "if continuous then differentable" is a false statement or it doesn't lead to any conclusions?
What was that ending? Vihart is taking over your channel?
you are the goat.
man this shows why i never got proofs. whenever i read them i tend to forget what i wanted to show in the first place or what the proof is supposed to show^^
Can you explain why f'=0 implies f is constant if the domain is the reals, but not if the domain is the p-adic numbers?
Simply explained
Great video as always
Great video. Thanks again.
Hello sir can I have a favor? do you lecture notes for Differentiability in R^2?
Can you help me with the following integral?
Integration of x^2 * √(1+4x^2) dx
It seems to me you can use integration by parts or something like trig substitution. However i get different results when doing so
Does this count as a real analysis question?
Oh man this came up in my analysis exam two months ago :0
Is it rigorous to say that the product of a derivative and an infinitesimal quantity is always 0?
integration of 1/x^2 + 3x ?
When you record a snail you find in your backyard
You are amaizing thank you
"I don't know what a number is..."
Are you sure you are the right person to make math videos?
@@christinabae2625 Thanks, anti-joke-chicken
@@christinabae2625 no problem. this comment is so old i don't even remember what i commented. But BPRP and I have some (friendly) jabs at each other from time to time.
Ok, to be honest it is me that does all the jabbing, but whatever
Clickbait: Table → Content. 😁
hm.. |0| = 0. But for f(x)=|x| df/dx at 0 don’t exist?
is there any other example of continuos functions that are not differentiable at some ponit?
Matheus Ramalho
For example f(x)=x^(1/3) is continuous but not differentiable at 0.
abs(x) is continuous but not differentiable at 0
I am here to blow your mind.
en.wikipedia.org/wiki/Weierstrass_function
Here's a nice little set of functions. f1(x) = sin(1/x).
f2(x) = x*sin(1/x).
f3(x) = x^2*sin(1/x).
Assume for each that the function is actually piece-wise defined where f(0) = 0 in order to "plug" the hole at 0.
f1(x) is not continuous at x=0 because the function oscillates between 1 and -1. f2(x) is continuous but not differentiable at x=0 because the tangent line slopes also oscillate.
f3(x) is differentiable at x=0 and every other point, however the derivative is not continuous, i.e. it is not second-differentiable.
Isn't a differentiable function by definition continuous?
Yes, and this is the proof.
Not by definition, the definition is that when you get closer to a certain x-value , the y-value you are approaching is the same as the value at the specific point of x. If that's true for all x then it's continuous. A differentiable function is a function for which all x-values the derivative is defined
thank ypu so much
Wait at the end it looks like the tangent formula
Will you discuss how the converse is not true and mention the Weierstrass function?
en.wikipedia.org/wiki/Weierstrass_function
What happened to the video where you found the limit of infinite square roots of 0? #yay
Alex Darcovich I will remake it
Nice
Lab glasses??
#yay!❤
I want a tshirt that says d'able
8:44 it’s me on math classes))))))
Can you solve this please?
X^lg(X-2)=1000
Тарас Заблоцький
Log as lg (base 10) or as ln (base e)?
NoName lg base 10 ,beacouse in spain we use log to say base 10
Does continuity also imply differentability?
HelloItsMe no
As explained in the video, a counterexample is f(x)=|x|. This is a continuous function, yet is not differentiable at x=0.
Weistrass function,
#yayyyyy
Sir, I don't get the proof..you just proved lim x → a f(x)=f(a)....where is the differentiability here?.. we are asked to prove if a f is differentiabie it is continous
You should've put another minute or two of that snail to get to 10 minutes. Do you not watch PewDiePie?