These things are really, really close, so let's say they're actually equall

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

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.

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)

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

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.

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

So if |f(x)−L|

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.

Even in the example OP gave: |f(x)−L|

holy fucking shit, is everyone in this thread retarded?
[math]|f(x)-L|

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

disagree. Now I'm merely an engineer brainlet so they might be right but the counter example I thought of here 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

retard
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]

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.

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

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.

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.

I kek'd.

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| 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.

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.

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

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.

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.

I'm a cs major. Pls no bully.

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