>these things are really, really close, so let's say

oly one of tehse can hold
=
>
<

You just don't understand even basics. sad fag noob

>pierdzioszek

>>8961936
I thought your critique was supposed to be rigorous. Is this a joke?

>>8961965
Most of theorems in real analysis basically say "if a and b differ at most by some small epsilon then we say they're equal", how is that not a meme?
>>8961946
and how does [math]|f(x)-L|<\varepsilon[/math] imply [math]f(x)=L[/math]?

>>8961989
>most
>basically
>>8961965

>>8961989
if they are not equal then there will be a fixed difference between them and hence an epsilon small enough that the difference between them is greater than epsilon

when there is no such epsilon it follows they must be equal. quite a simple concept, so I don't think analysis is for you.

File: topology.jpg (35KB, 711x388px) Image search: [Google]

35KB, 711x388px

>>8961936
>Topology tell us that
a Mug is equal to a Donut (Torus)

File: just_call_me_torus_blue_torus_topology_coffee_mug-r8ff4aa5a8071483c8ef3ee74488ac484_x7jgr_8byvr_540.jpg (23KB, 540x540px) Image search: [Google]

23KB, 540x540px

>>8961936

a Mug is equal to a Donut (Torus)

t. Topology

>>8961989
>it's another "brainlet criticizes things he doesn't understand" episode

>>8961989
Becouse it holds for all epsilon in positive reals. Say f(x) = L+q where q is an arbitary real. Then |f(x)-L|=|q|. But if q isnt zero, we could put epsilon=|q/2| and the inequality wouldn't hold. Therefore q is zero and f(x)=L.

>>8962007
>>8962189
I thought it only meant that the limit of f(x) approached L.

(I haven't done real analysis, only epsilon-delta limit definition in calc)

>>8962189
so why not just say "a and b are equal" instead of this epsilon bullshit?

>>8962651
That's how it is often taught,unfortunately. There's a lot of misconceptions about the relationship between sequences and limits. There's a lot of misconceptions about functions and limits too.

That's why we have real analysis. It is a rigorous construction of what we can and can't say about the number line, and n dimensional spaces in general.

In order to be rigorous, it needs to be complete,unambiguous and self consistent. Youre just at a point where you feel your logic is valid and that it is being contradicted. If something doesn't make sense, then most likely, your sense is wrong.

>>8962659
because functions and sequences are not numbers. Just because the sequence or function contains the number isn't sufficient to show that that number is actually the limit. At least not always

>>8962666
So if |f(x)−L|<ε then f(x) we can say that f(x) is actually equal to L, not just it's limit approaches L?

In that case what happens at point discontinuities where L is defined but f(x) isn't?

>>8962651
>>8962659
The statement is different in limits, which states for all epsilon in positive reals, there exists delta in positive reals so that...

It can be that the limit doesn't equal the function value.

Read Rudin's Principles of Mathematical Analysis, it all comes clear then.

>>8962681
Even in the example OP gave: |f(x)−L|<ε

What if we define a function f(x) = 9 at x=0 and x^2 everywhere else, and L(x) = x^2 everywhere.

Then since epsilon is always positive, even though they will be equal to each other within the neighborhood around 0, at 0 they won't be equal.

holy fucking shit, is everyone in this thread retarded?
[math]|f(x)-L|<\varepsilon\;\;\forall \varepsilon[/math] does not fucking imply [math]f(x)=L[/math], it implies [math]\lim f(x)=L[/math]. It is a definition for the limit, it does not imply in the slightest that [math]f=L[/math], but instead, that it is sufficiently close.

>>8962748
That's what I've been trying to say but

>>8962681
>>8962666

disagree. Now I'm merely an engineer brainlet so they might be right but the counter example I thought of here >>8962722 seems to contradict what they say.

Equivalence relations other than literal equality are a fucking joke. You're basically saying "oh these things are equivalent so they're basically equal", fuck you

>>8962787
retard
>>8962756
intuitively, a function is continuous if [math]\lim_{x\rightarrow c} f(x)=f(c)[/math]. In the "counter example", if you will find that for [math]x[/math] close enough to the discontinuity, [math]|f(x)-L|\nless \epsilon[/math], for some [math]\epsilon[/math], say, [math]\epsilon=8[/math]

>>8962679
The dude you're responding to has no idea what he's talking about. Epsilon-delta gymnastics talk about limits and continuity, this has nothing to do with f(x) = L. f(x) either is defined on some given x_o and it has some value that either does or doesn't equal L. just because we can say that if x lies in the delta neighbourhood of x_o then f(x) lies in the epsilon neigbourhood of L doesn't mean that the function actually equals L at x_o. That dude is wrong.

>>8962787
have you not even studied congruences? are you in high school?

>>8962815
What exactly did I say that was wrong?
If the function has a limit that does not diverge, it is continuous. You do realize that extreme value, intermediate value, and lipschitz and cauchy definitions that are used for looking at functions are built directly from delta epsilon proofs of sequences right?

All I said is that f (x) = L at c isn't sufficient by itself to prove the limit exists, which you are in agreement with. If more is known about the function, it can be true but not necessarily, which is why we need the proofs in the first place.

>>8962840
oh so now two triangles are the same just because they're congruent? What if they're 100 million miles apart and one of them is upside down?

Congruence is retarded.

>>8962845
I kek'd.

>>8962809
Ah i think I got it now. You can disprove my counterexample since at x=0, |f(0)-L| = 9 > ε, hence we cannot say the two functions are equal.

It's pretty much the generalized version of

|a-b|<ε => a=b

if we say that the above has to hold with f(x) and L(x) for all values of x in the domain of both.

>>8962867
No. L is a number, not a function.

It also doesn't apply for all x in the domain because the limit can be different at each x.

>>8962722
L is a constant. Hell are you all high schoolers? This all is taught on Analysis I.

>>8962748
Holy hell, I just fucking gave a proof that clearly shows f(x)=L in this case. Limits need delta epsilon. Mere epsilon shows equality.

>>8962953
Oh. Then why don't you write f(a) sempai? To me f(x) standards for the function itself. If you want to show the function evaluated at a particular value than I think f(a) is clearer.

File: file.png (99KB, 323x250px) Image search: [Google]

99KB, 323x250px

>>8962988
I'm a cs major. Pls no bully.

>>8963019
It's just convention to say lim f (x) as x->c
Most of the time, you can just plug in c but for example,

f (x) = cos (1/x) is not defined at 0 but the lim f (x) as x->0 = 1