I wish my teacher taught that like this way. He just goes too fast that we cannot follow what he's doing xd. Thanks a lot Khan Academy!
@giack6235Ай бұрын
Hello and thank you for the video, but I think there's an error in 9:44: If you have lim (x -> c) (f(x) - f(c)) = 0, you can conclude that this is equivalent to: lim(x -> c) f(x) = lim(x -> c) f(c) only if you know that the single limits exist, but you don't know yet if lim(x -> c) f(x) actually exist!
@cansucorbac15145 жыл бұрын
"The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn." :)))) Canım.
@MyAsdfqwe5 жыл бұрын
canın.
@esraeren003 жыл бұрын
actually calculus course really require knowling the proof! ın the exams 75 percent of questions are measued our ability to prove them
@booguy26362 жыл бұрын
Why do you start the proof with lim x→c (f(x) - f(c))?
@TU7OV6 ай бұрын
Because to show that lim x->c f(x) = f(c) you have to show that lim x->c f(x) - lim x->c f(c) =0 or lim x->c (f(x) - f(c)) = 0, this follows from the limit laws.
@aanyaakhandelwal17784 жыл бұрын
Thank you very much!!! So good!!!
@mehulkathiriya17378 жыл бұрын
awesome...I like it even more than iit lecture.short and simple....
@serve932 Жыл бұрын
I need to learn this 4 times for it to get inside my head
@benwinstanleymusic3 жыл бұрын
Thank you very much Sir!
@arunkumar4u4 жыл бұрын
Why you took lim( f(x)-f(c)) ?? x>c
@booguy26362 жыл бұрын
I'm wondering the same thing.
@SadikulIslam-cm8wo3 жыл бұрын
Really awesome
@rajeshghosh87378 жыл бұрын
Nice explanation.........thanks.
@Hakahah_o Жыл бұрын
Thank you
@jamessamuel12554 жыл бұрын
Bloody brilliant
@cyvex12814 жыл бұрын
Short and simple
@rainerwahnsinn32626 жыл бұрын
You are only allowed to split the limit, if both parts are *convergent*!
@manivannan235 Жыл бұрын
Yes it is indeed convergent😊
@stickmanbattle9972 жыл бұрын
This video only proves limx->c(f(x)) = f(c), but the most important info if the lim is not continuous at that point it means is not differentiable.
@sujitbaruah45365 жыл бұрын
Nice lecture
@wagsman99995 жыл бұрын
Excellent. I suppose you can assume f(c) exists, because if it didn't the derivative would fail.
@Lark-um8hv Жыл бұрын
Can't have the slope at a point if the point doesn't exist
@ghassanayyad97669 жыл бұрын
thank you so much
@aanyaakhandelwal17784 жыл бұрын
Yes, you are very corrrect!
@randomeggthatworksforthefb71724 жыл бұрын
Plus eggs are tasty
@hassanakhtar78744 жыл бұрын
Œ
@lizmurith98618 жыл бұрын
Why do you multiply and divide by (x-c)? Why are we changing the original formula?
@MrCanadianplayer8 жыл бұрын
Because it's a constant Liz! (x-c) over (x-c) is just like saying 1/1=1
@lizmurith98618 жыл бұрын
Manchester United Videos Thanks!!!
@MrCanadianplayer8 жыл бұрын
Liz Menezes no problem lolll
@MrCanadianplayer8 жыл бұрын
Liz Menezes I actually have a test on this stuff tomorrow haha can you answer my question now? Why is Lim x approaching c of x-c =0? Hope you can help me out Liz! Matan
@MrCanadianplayer8 жыл бұрын
Liz Menezes so F(c) =0?
@jaswanthramesh24013 жыл бұрын
can f'(C) be infinity?(happens for a vertical straight line) , if it is infinity, then 0*infinity becomes an indeterminate form . Then we wont get this.
@jakubsmyk22513 жыл бұрын
if we assume differentiability of function f at point C then f'(C) is finite
@akames.529 Жыл бұрын
7:24 where does (x-c) come from?
@Lark-um8hv Жыл бұрын
w = (q/q)w = (w*w)/q = w × (w/q) Try it with constants
@DynV Жыл бұрын
TLDR 10:56
@CMDF084 жыл бұрын
I think I missed something here. Why can we say that the limit of f(c) as x approaches c is equal to f(c)? I don't see where we proved that anywhere. I don't understand where it comes from. I must have missed some detail. I understand how we proved that the limit of f(x)-f(c) as x approaches c = 0, but if our proof is using the fact that f is continuous at c, isn't that circular logic? Or does the limit of f(x)-f(c) as x approaches c = 0 somehow require f to be continuous at c? I'm so lost :(
@jeanbadgenes84604 жыл бұрын
yeah, he's wrong, they are not equal, thats why is discontinuous
@tansu14993 жыл бұрын
He's right because f(c) is a number. that's one of the rules of limit , when you want to find limit of a number the answer "always" is that number. so here limit of f(c) always equals to f(c). If you didn't get it just watch a video about rules of finding limit.
@CMDF083 жыл бұрын
@@tansu1499 I think I understand now. Since f(c) evaluates to the value at c regardless of whether the function is continuous or not, then f(c) would certainly be some unknown number. My only issue now is what if f(x) were undefined at c? In that case, we could not say f(c) is a number, right? Aren't we also depending on the idea that a function that is differentiable at c is also defined at c? Perhaps there is another, independent proof that establishes this though, that I've forgotten as I haven't looked back at this for a while.
@CMDF083 жыл бұрын
Now that I look at the formula for differentiation, actually, perhaps it is trivial to say that if f(c) were undefined at c. We would not be able to apply that formula otherwise, and therefore it must be defined at c if it is differentiable at c.
@tansu14993 жыл бұрын
@@CMDF08 exactly. We know that it's differentiable at x=c so it is defined at this point and then we use these steps to prove that it is also continuous at this point.
@Dyslexic-Artist-Theory-on-Time9 жыл бұрын
Can we think of this as a process over a period of time?
@Lark-um8hv Жыл бұрын
Sure, if x represents time why not?
@drallisimo349 жыл бұрын
cool!!!
@aragonification8 жыл бұрын
But if differentiability at a point means that the derivative of that point exists, why the first example from the right side is not continuous?
@wolframalpha86346 жыл бұрын
As per the definition of continuity lim as x approaches point c from the right as well as the left and also if the function is defined at that point C then it is equal to f(c), whereas the function at point c is defined at some other value of f(x) , i mean a removable discontinuity... instead if we were to look at the actual position of f(c)(marked with an empty circle) the left hand limit and the right hand limit exists at that point but is not equal to f(c) cause it is not defined at that point!! Hence the function is neither continuous nor differentiable at that point
@imdoug9 жыл бұрын
math god
@puspitapaul7257 жыл бұрын
please explain the differentiability of |log|x||
@renayachan57147 жыл бұрын
Does it mean that if a function aint continuous at a number it is also not differentiable at that number?
@montrast07057 жыл бұрын
Yes. A implies B is equvialent to notB implies notA. In your case A being function is differentiable and B being function is continuous.
@Mooreeezy7 жыл бұрын
well differentiability implies continuity. however continuity does not necessarily imply differentiability. Which means that if the function is not continuous then that function could sill be differentiable.Because differentiation is basically the process of finding the derivative.And a derivative is a means whereby we are able to find the slope of a non linear function at any point on that function. Continuity suggests that a function is continuous. Does a function need to be continuous in order to be able to use the same derivative to find any slope on it? I don't believe so. So, in my mind that means that continuity does not imply differentiation. which I think means a function does not have to be continuous to be differentiable. *UPDATE* thought about it some more. If you have a gap and you use the derivative to find the slope of the function with the gap, if you plug an x value into the derivative and it ends up in the gap, that would suggest that the derivative does not work for that specific point on the graph. which means you would have to use a different derivative. which means the graph wouldn't be differentiable at that number. So I guess I really don't know.
@Muhammed_English3144 жыл бұрын
@@Mooreeezy If you know a little bit about logic (like me) you will conclude that the answer is yes because "discontinuity implies undifferentiability" is the contrapositive of "differentiability implies continuity"
@vvr3117 жыл бұрын
what can you say about differentiability of y = x^(1/3) and y = x^(2/3)?????
@Muhammed_English3144 жыл бұрын
they are both differentiable on the real numbers except at x=0 1) y'=1/(3cube root of x squared) so if you look at the domain of the derivative function it's R/{0} 2) just the same but without the square
@majorgeneralrahul62983 жыл бұрын
Fun Fact :- Even if you assume it's NOT differentiable you'll get the same answer if you use this method. :D
@kamitube10594 жыл бұрын
I lost at 7:10 , can anybody please help me? why is he multiplying (x-c) at top and bottom?
@lebongarcon33904 жыл бұрын
he multiplies it top and bottom so it is equal to the original limit function. He is doing this to connect the derivative function to the limit function
@paulpandig93255 жыл бұрын
try to arrange the video in order that means first proof and then sum
@yashuppot32145 жыл бұрын
The audio...
@ian.ambrose3 жыл бұрын
I know. It's great.
@ahmedalsherbini2768 жыл бұрын
thumb up
@prajwol_poudel4 жыл бұрын
writing this in my exams aint gonna score me good marks
@Muhammed_English3144 жыл бұрын
learning is not for grades it's for science and fun.