Real Analysis | Sequential limits in functions.

We prove a nice result that allows us to use technique of limits of sequences to approach limits of functions.

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[stabbysmurf]

I've always wondered why we don't just use the sequential definition as the primary one. Sequences are such an effective workhorse for establishing fundamental results in real analysis, you might as well use it here as well. And it is more intuitive to say "no matter how you approach a, f approaches L" than the epsilon-delta cha-cha.

I guess one reason for the usual definition is that we can define uniform convergence in a very similar way.

[oussamadjeziri]

been looking for this all day. REALLY, Thank you.

[backyard]

the last problem was pretty intuitive. the limit as x goes to zero of sin(1/x) means that the argument of sine tends towards infinity as x goes to zero, but limit as x goes to infinity of sin(x) obv doesnt exist due to sine just oscillating, i.e. if you picked L to be any real number you could show easily using epsilon delta that it isnt a limit.

[debendragurung]

The theorem looks neat. But again the proof just flew over my head.
I think this is more concrete definition of the limit of the function.

[ashwinvishwakarma]

Just a slight subtlety that you had to show x_n=/=a for any n. But that comes from the definition of the limit where you say |x_n - a|>0, so x_n=/=a for any n (also because x_n is chosen from A and a is a limit point of A).

[danwe]

Wow! Heine's theorem! I haven't seen this boi for ages! :O :D

[사기꾼진우야내가죽여]

Since we have proved that sequential limit of function is equivalent to the original definition of limit of function, we can use this fact to show extreme value theorem.

[truonghuynhvan]

The video sound is pretty good, beyond my imagination

[autishd]

I wonder whether you could help me solve question 23 from 2020 AMC 10B. The question is as follows: Square ABCD in the coordinate plane has vertices at the points A(1, 1), B(-1, 1), C(-1, -1), and D(1.-1). Consider the following four transformations:
- L a rotation of 90 degrees counterclockwise around the origin;
- R, a rotation of 90 degrees clockwise around the origin;
- H, a reflection across the x-axis;
- V, a reflection across the y-axis.

Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. For example, applying R and then V would send vertex A at (1, 1) to (-1, -1) and would send vertex B at )-1.1) to itself. How many sequences of 20 transformations chosen from {L, R, H, V} will send all of the vertices back to their original positions?

(a) 2^37
(b) 3.2^36
(c) 2^38
(d) 2^39

[anshumanagrawal]

You need 0<|x-a| in your definition

[schweinmachtbree]

definition of the limit of a function is wrong. should be 0 < |x-a| < delta instead of |x-a| < delta ...

[franksaved]

In Italian analysis courses they call it the Bridge theorem, wtf.

[alexwestworth]

Notice how Michael structures videos similar to, and speaks using the formal language of, mathematics journal papers. And that's a good place to stop.

[orenfivel]

would like to show a contra positive of ⇒
(∃{a_n} a_n→a ∧ f(a_n)↛L) ⇒ lim{f(x), x→a}≠L

PF:
foil out the definitions for each statemet in the "and" operation (∧) above:
∃{a_n}:
∀δ>0 ∃N(δ)∈ℕ ∀n∈ℕ, n> N(δ)⇒|a_n-a|<δ; and
∃ε>0 ∀M∈ℕ ∃n∈ℕ, n>M ∧ |f(x)-L|≥ ε.

Set M= N(δ), find n> N(δ) s.t |a_n-a|<δ (b/c a_n→a) but |f(a_n)-L|≥ ε (b/c f(a_n)↛L). hence
∃ε>0 ∀δ>0 ∃x∈ℝ (x=a_n for some n> N(δ)), |x-a|<δ ∧ |f(x)-L|≥ ε.
⇒ lim{f(x), x→a}≠L ∎