I remember my first Calculus I classes, where this formal definition appeared and the entire class felt completely overwhelmed by the professors explanation. Most of the time, things are simpler than we think and, as teachers, we should be able to simplify it.
by far, this video is undoubtedly the best intuitive explanation about the formal definition of a limit. It has everything I love about maths , Constructing the formal ideas based on intuitive concepts that everybody has in their minds, and connecting these ideas in order to produce valuable mathematical concepts
1. lim(x->a+)f(x)=L For all epsilon greater than 0 there exists a delta greater than 0 such that 0< (x-a)< delta and |f(x)-L|< epsilon 2. lim(x->a-)f(x)=L For all epsilon greater than 0 there exists a delta greater than 0 such that 0< -(x-a) < delta and |f(x)-L|< epsilon 3 and 4 take more effort and I have finals in a week :P I should sleep, I learned to type in the dark on this keyboard finally haha.
A good way to practice is trying to think why lim f(x) = sin(pi/2x) fails to comply with this definition. Again, thank you for this explanation, good sir.
Would Q3 be… For every delta > 0, there exists a real number M such that: If 0 < | x - a | < delta, then f(x) > M Is that correct…? Sorry am new to this!
hmmm, let's say that we have f(x) = 3. (For any a in the domain) lim x->a f(x) = 3. M can be any real number, take M= 4. For all positive delta, if 0 < | x - a | < delta, then 3 = f(x) > M = 4 So I'm unsure. I'm stuck on this too (though I'm assuming that you've already graduated from this course so if you've got some godly insight now that you're more knowledgeable, I'd love to hear it because I'm stuck on this one ;u;)
if x is within delta of the limit point , then f(x) will be within epsilon of L. f(x) CANNOT be outside epsilon, if it is, the limit does not exist at L
Exercise 1solution Let a+, L € R and let f € I which is at least centred at a+ except maybe at a+, then we say that limf(x) as x approaches a+ = L, when for all £>0, there exist d > 0 s.t for all x € R. 0
Why do we still use the absolute value even though we know we're approaching x from the positive direction? I wrote mine as " for all £>0, there exists d>0, s.t (0 < x-a < d) implies f(x) - L < £ "
@@umarfaruk328 In that case we were working with limits from both directions. So the absolute value is used for the distance whether being negative(x < a) or positive (x > a). But in this case we know that it's approaching a from the positive direction meaning x > a so I don't see the use for the absolute value sign.
@ George , one of the implications of removing the absolute sign will be for example |f(x)-L| f(x) is between L+£ and L-£, which is derived from the algebraic definition of absolute values. So what do u think will happen if u remove the absolute sign, would it still be the same?, I don't think so.
@@umarfaruk328 That means there's no real difference between the definition of the limit from both directions (normal scenario) from the definition of a side limit except the addition of the sign? I think |F(x) - l| < £ should remain because the value of F(x) can vary depending on the function, but |x - d| < 0 should not because we already know which value is larger between (x - a) and d.
"no matter how close to L epsilon might get, we will always be able to find a delta such that for every x in between delta and a, f(x) is always in between epsilon and L"
Exercise 4solution Let a, infinity € R and let f € I which is at least centred at a except maybe at a, then we say that limf(x) as x approaches a = - infinity, when for all £>0, there exist d > 0 s.t for all x € R. 0
I remember my first Calculus I classes, where this formal definition appeared and the entire class felt completely overwhelmed by the professors explanation. Most of the time, things are simpler than we think and, as teachers, we should be able to simplify it.
by far, this video is undoubtedly the best intuitive explanation about the formal definition of a limit. It has everything I love about maths , Constructing the formal ideas based on intuitive concepts that everybody has in their minds, and connecting these ideas in order to produce valuable mathematical concepts
Finally I have found a video that precisely describes each part of the definition. Thanks a lot!
The bit about the hidden quantifier cleared up all my doubts about this. Thanks.
A very intuitive explanation that greatly aided my comprehension. Thank you sir!
Hello ,thank you this is the best video about the formal definition limit on youtube.
There are many videos that 'tries' to explain the formal definition of limit. This video did it best.
1. lim(x->a+)f(x)=L
For all epsilon greater than 0 there exists a delta greater than 0 such that
0< (x-a)< delta and |f(x)-L|< epsilon
2. lim(x->a-)f(x)=L
For all epsilon greater than 0 there exists a delta greater than 0 such that
0< -(x-a) < delta and |f(x)-L|< epsilon
3 and 4 take more effort and I have finals in a week :P I should sleep, I learned to type in the dark on this keyboard finally haha.
for Q3 and Q4 we could say:
Q3: ∀M>0,∃δ>0 s.t. 00,∃δ>0 s.t. 0
God-tier video. Thank you sir.
Explanations are very clear and easy to understand!!! Thanks so much!!!
Never knew the importance of this when learnt for the first time
Yeeeeesssssss. Thank you!!!!! This is exactly what i was looking for
A good way to practice is trying to think why lim f(x) = sin(pi/2x) fails to comply with this definition.
Again, thank you for this explanation, good sir.
Thanks a lot
just pure beauty
Would Q3 be…
For every delta > 0, there exists a real number M such that:
If 0 < | x - a | < delta, then f(x) > M
Is that correct…? Sorry am new to this!
hmmm, let's say that we have f(x) = 3. (For any a in the domain) lim x->a f(x) = 3. M can be any real number, take M= 4. For all positive delta, if 0 < | x - a | < delta, then 3 = f(x) > M = 4
So I'm unsure. I'm stuck on this too (though I'm assuming that you've already graduated from this course so if you've got some godly insight now that you're more knowledgeable, I'd love to hear it because I'm stuck on this one ;u;)
@@dxtura mate, you can't choose f(x) = 3 then for all a limit is equal to 3.
@@duetothefacthat yea I’ve noticed that since my comment lmao, thx for the input though!
This is wonderful! Thank you
best explanation
very useful
Superb👍
Great explanation but should the definition not be for every Epsilon there must exist ONE delta greater than 0?
Isn't that what the definition says?
I'm lost at the forbidden constraint part what does it mean? from 3:56
if x is within delta of the limit point , then f(x) will be within epsilon of L. f(x) CANNOT be outside epsilon, if it is, the limit does not exist at L
I have no idea how to do the exercise given at the end of the video.. guess I don't understand after all :(
If you're having trouble understanding the concepts from the videos alone, feel free to come to office hours or to ask on Piazza!
geez me too I still don't understand the concept and can only memorize it
It might help to watch 2.6! I was stuck until I watched the next video, then it all clicked.
Exercise 1solution
Let a+, L € R and let f € I which is at least centred at a+ except maybe at a+, then we say that limf(x) as x approaches a+ = L, when for all £>0, there exist d > 0 s.t for all x € R.
0
Why do we still use the absolute value even though we know we're approaching x from the positive direction? I wrote mine as " for all £>0, there exists d>0, s.t (0 < x-a < d) implies f(x) - L < £ "
Don't forget we are working with the definition and the definition has absolute sign also
@@umarfaruk328 In that case we were working with limits from both directions. So the absolute value is used for the distance whether being negative(x < a) or positive (x > a). But in this case we know that it's approaching a from the positive direction meaning x > a so I don't see the use for the absolute value sign.
@ George , one of the implications of removing the absolute sign will be for example |f(x)-L| f(x) is between L+£ and L-£, which is derived from the algebraic definition of absolute values. So what do u think will happen if u remove the absolute sign, would it still be the same?, I don't think so.
@@umarfaruk328 That means there's no real difference between the definition of the limit from both directions (normal scenario) from the definition of a side limit except the addition of the sign? I think |F(x) - l| < £ should remain because the value of F(x) can vary depending on the function, but |x - d| < 0 should not because we already know which value is larger between (x - a) and d.
"no matter how close to L epsilon might get, we will always be able to find a delta such that for every x in between delta and a, f(x) is always in between epsilon and L"
Exercise 4solution
Let a, infinity € R and let f € I which is at least centred at a except maybe at a, then we say that limf(x) as x approaches a = - infinity, when for all £>0, there exist d > 0 s.t for all x € R.
0
Surely infinity is not in the set of reals?
everybody, dont listen to this guy
:)