How to prove a limit exists and that a function is continuous at a point?

[whole sun]

I'm a little confused by what the epsilon-delta definition is used for. So far, I've been proving that a limit exists by approaching it from the left and right, and showing that they're equal. So where does epsilon-delta come in?

I have a homework question asking me to prove that some function is continuous at some point, so the way I'm planning to do this is by proving all 3 requirements of continuity hold at said point. One of these requirements is to show that the limit exists as x approaches the point. I don't think I can use the approach from left and approach from right method here, because that usually uses relies on the function being continuous so we can do this $lim_{x\toa}f(x)=f(a)$.

So should I use epsilon-delta for this? If so, how?

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[warm trellis]

the way you usually do it

[bronze saffron]

ElectricTortoise
Compile Error! Click the reaction for more information.
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[warm trellis]

the definition of limit is not dependent on continuity

[modern garnet]

Can you supply us with the exact question so we can help you more specifically

[whole sun]

I am aware of that

part a

[nova heron]

The epsilon-delta definition is used for, y'know, defining the concept of a limit.

Like, every time you ever talk about a limit, the epsilon-delta definition is what you're actually saying.

[whole sun]

I see

Can it be used to prove the existence of a limit?

Or is that done by other means

[nova heron]

I mean, it must be.

[warm trellis]

the limit does exist

[whole sun]

How do I prove that

[nova heron]

Because otherwise saying "the limit exists" is literally meaningless.

[modern garnet]

The existence of a limit is a statement that really has meaning only in terms of the delta elsilon

[warm trellis]

pick delta = epsilon for small epsilon

$$|x\sin (1/x)| \leqslant |x| < \delta$$

[bronze saffron]

aL

[nova heron]

That said, if you're allowed to not directly reference the epsilon-delta definition, this is not that difficult to prove.

Like the question says, use the squeeze theorem.

[whole sun]

I suspected as much

Our profs don't plan on teaching epsilon-delta unfortunately

So I have to go out of my way to learn it myself

[warm trellis]

this is sufficient, you can squeeze for higher powers of x

[nova heron]

Eh, it's more of a backend type deal.

[whole sun]

Yes ik

We are an engineering university

-my prof

[nova heron]

You only actually use it to prove a bunch of properties of limits, and then you use those properties and never look at the definition again.

[warm trellis]

dios mio :/

[modern garnet]

A very big bunch of properties...

[whole sun]

like the sequential criterion thingie

[nova heron]

Or like the Dedekind construction of the reals.

[whole sun]

that was genuinely so cool, but I didn't understand the proof

[warm trellis]

indeed, sequences are much easier to work with than going epsilon chasing

[nova heron]

All you need that for is to prove that a complete ordered field exists, and then once you have that you can prove that all complete ordered fields are isomorphic to one another, and then you can stop caring about how specifically the reals are constructed.

[whole sun]

That is something I might look into if I have the time

[modern garnet]

Category theory mindset

[whole sun]

uni is hard 😢

[warm trellis]

learn the properties of limits and squeezing technique(s)

that'll be sufficient

[whole sun]

I was looking at the proof that someone on this server gave me (think it was aL I'm not sure) and I was wondering what this doohickey is

[whole sun]

well yes ofc it will be otherwise the profs wouldn't be teaching it

but I want more than just sufficient

[warm trellis]

it's the open delta ball around the point c

$$U_\delta (a) = {x \colon |x-a| < \delta}$$

[bronze saffron]

aL

[whole sun]

do we need 0 < |x-a| also?

[nova heron]

$|x - a| < \delta$? Not $0 < |x - a| < \delta$?

[bronze saffron]

Techie Literate

[whole sun]

I was watching bprp explain epsilon-delta and one thing he puts emphasis on is the fact that we don't actually care what happends at x=a

[nova heron]

If you just want an intuitive understanding of the definition, I can explain.

[warm trellis]

that's already baked in the proof earlier

[nova heron]

I don't see it. The first line of the proof is in the screenshot, and V_delta is never defined, just invoked.

[whole sun]

I've watched bprp and 3b1b explain the intuitive meaning of the definition, but I think I need to practice to really get the hang of it

[nova heron]

I mean. I think I might be able to do better.

Or at least give you another angle on it.

[whole sun]

ooo sure why not

[nova heron]

So the basic idea of a limit is this. When we say $\lim_{x \to a} f(x) = L$ what we're actually saying is that we can get f as close as we like to L by getting x close enough to a.

[bronze saffron]

Techie Literate

[warm trellis]

yes you do, passive learning will only carry you so far

[nova heron]

So the definition is constructed just to capture this idea.

Epsilon is how close we want f and L to be, and delta is how close x and a have to be to make that happen.

So to be able to get f as close as we like to L no matter how close that is, we require that for all positive epsilon, a delta exists that gets f within epsilon of L.

[warm trellis]

this lingo is meant to fix all the "infinitely close, infinitesimal" and all that nonsense you hear about in integration

[nova heron]

That's not even how hyperreals work anyway.

[whole sun]

3b1b did mention that

"infinitely small is a nonsensical term"

[nova heron]

I mean, that's nonsensical for two reasons.

[whole sun]

well "inifinitely" is one of them

what's the other

[nova heron]

It's nonsensical to talk about an amount of "smallness".

Whether that's an infinite or finite amount.

"This apple is three units small." What does that statement mean?

[whole sun]

mmmmmmmmm alright

[nova heron]

It makes sense to talk about something being "infinitely large", both in a limit context where you mean simply that it grows without bound, and in the sense of literally having an infinite amount of something, e.g. the set of natural numbers is infinite.

[whole sun]

hmm I never noticed that

"infinitely large" makes sense, but "infinitely small" doesn't

that's pretty neat

matches up nicely with the concept of infinity too

[nova heron]

If you want to get technical, I guess "infinitely small" would just refer to a negative infinity.

Except the phrase is usually specifically used to address the concept of infinitesimals.

But the definition of an infinitesimal is actually much more straightforward.

e is infinitesimal iff 0 < e and for all x in R, 0 < x implies e < x.

[whole sun]

ooo

[nova heron]

It's the "smallest positive number", which is different from being "infinitely small".

[whole sun]

whoa but that's kinda nutty

[nova heron]

And impossible in the real numbers.

[whole sun]

so e is not in R

[nova heron]

Right, because the reals are dense.

That is, between any two real numbers is a third real number.

[whole sun]

because otherwise we could have x=e

[nova heron]

If e was in R, then, because 0 is in R, by the density of the reals we would therefore necessarily have an x in R such that 0 < x < e.

[whole sun]

this is knowledge I forfeited when I came to the engineering university instead of the research one 🥀

[whole sun]

whoa that's cool

[whole sun]

I can try to learn it myself but it won't be the same

[nova heron]

We say that a set S ordered by an ordering relation < is dense iff, for all a, b in S such that a < b, there exists c in S such that a < c < b.

Examples of dense sets are the reals and the rationals.

Non-dense sets include the naturals and the integers.

[whole sun]

mmm ok

[nova heron]

Well, the naturals and integers aren't dense given the standard ordering relation <, but you could define an ordering relation over which they would be dense.

[whole sun]

do we have a name for the set R U e

[nova heron]

Probably.

[whole sun]

the reals + infinitesimal

[nova heron]

R U {infinitesimal, infinity} is known as the hyperreal number system.

[whole sun]

oooooo ok

so that's what you were talking about

[whole sun]

Can squeeze theorem be used to show a limit exists?

I thought it was used to find what the limit converges to

[warm trellis]

yes

[warm trellis]

you can also determine the limit by squeezing

[whole sun]

So if I can squeeze a limit, it exists?

[warm trellis]

this is squeezing for example

|x| -> 0 as x to 0, hence the to-be-squeezed quantity also converges to 0

that's the whole point of squeezing, to determine limits that are difficult to calculate directly

[nova heron]

Do you recall what the statement of the squeeze theorem is?

[whole sun]

If $h(x) \leq f(x) \leq g(x)$ and $\lim_{x\to a}h(x) = $\lim_{x\to a}g(x) = L$ then $f(x) = L$?

[bronze saffron]

ElectricTortoise
Compile Error! Click the reaction for more information.
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[nova heron]

Well, almost. So if $\lim_{x \to a} f(x) = L$, then it obviously exists, right?

[bronze saffron]

Techie Literate

[whole sun]

Hmm ig that makes sense

[nova heron]

And actually the proof of the squeeze theorem uses the epsilon-delta definition of a limit.

[whole sun]

So what you're saying is that when a limit doesn't exist, I cannot squeeze the function

Ooo ok

I'll look into that too 👍

[nova heron]

I can actually present the proof.

[whole sun]

Ooo ok

[nova heron]

So we know $\lim_{x \to a} h(x) = L$, right?

[bronze saffron]

Techie Literate

[nova heron]

That means that, for all $\epsilon_h > 0$, there exists $\delta_h > 0$ such that, for all real x, $0 < |x - a| < \delta_h$ implies $|h(x) - L| < \epsilon_h$.

[bronze saffron]

Techie Literate

[nova heron]

We also know that $\lim_{x \to a} g(x) = L$, therefore for all $\epsilon_g > 0$ there exists $\delta_g > 0$ such that $0 < |x - a| < \delta_g$ implies $|g(x) - L| < \epsilon_g$.

[bronze saffron]

Techie Literate

[nova heron]

We wish to prove that, for all $\epsilon_f > 0$, there exists some $\delta_f > 0$ such that $0 < |x - a| < \delta_f$ implies that $|f(x) - L| < \epsilon_f$.

[bronze saffron]

Techie Literate

[nova heron]

Now, let's observe these inequalities.

We have $h(x) \leq f(x) \leq g(x)$.

[bronze saffron]

Techie Literate

[nova heron]

For x within some neighborhood of a.

Now, what does $|h(x) - L| < \epsilon_h$ mean?

[bronze saffron]

Techie Literate

[nova heron]

Well, it means $-\epsilon_h < h(x) - L < \epsilon_h$, or that $L - \epsilon_h < h(x) < L + \epsilon_h$ for all x such that $0 < |x - a| < \delta_h$.

[bronze saffron]

Techie Literate

[nova heron]

Or, for all x such that $x \neq a$ and $a - \delta_h < x < a + \delta_h$.

[bronze saffron]

Techie Literate

[nova heron]

This should maybe hammer in a bit the intuition about "closeness".

Similarly, for all $x \neq a$ such that $a - \delta_g < x < a + \delta_g$, we have $L - \epsilon_g < g(x) < L + \epsilon_g$.

[bronze saffron]

Techie Literate

[nova heron]

Thus, if we let $\delta_f = \min(\delta_g, \delta_h)$, then we get $L - \epsilon_h < h(x) \leq f(x) \leq g(x) < L + \epsilon_g$.

[bronze saffron]

Techie Literate

[nova heron]

Thus $\epsilon_f = \max(\epsilon_g, \epsilon_h)$.

[bronze saffron]

Techie Literate

[nova heron]

Or, putting it another way, given $\epsilon_f$, let $\epsilon_g = \epsilon_h = \epsilon_f$.

[bronze saffron]

Techie Literate

[nova heron]

Then choose $\delta_f = \min(\delta_g, \delta_h)$.

[bronze saffron]

Techie Literate

[nova heron]

Did you understand?

The corollary here is that if $\lim_{x \to a} f(x)$ doesn't exist, then there do not exist $h(x)$ and $g(x)$ such that $\lim_{x \to a} h(x) = \lim_{x \to a} g(x) = L$ and $h(x) \leq f(x) \leq g(x)$ within a neighborhood of a.

[bronze saffron]

Techie Literate

[nova heron]

Either f must escape the bounds of g and h or the limits of g and h must not equate.

[whole sun]

how did you go from $\epsilon_f = max(\epsilon_g, \epsilon_h)$ to $\epsilon_g = \epsilon_h = \epsilon_f$.

[bronze saffron]

ElectricTortoise

[nova heron]

I was wrong. We get to choose e_g and e_h, so choose them to equal e_f.

[whole sun]

Oh so that implies $\delta_f = min(\delta_g, \delta_h)$

[bronze saffron]

ElectricTortoise

[nova heron]

What I meant to say is when 0 < |x - a| < min(d_h, d_g), we get that extended inequality.

[whole sun]

This one right

Ohhh ok I get it

Because $\epsilon_f = \epsilon_g = \epsilon_h$

[bronze saffron]

ElectricTortoise

[whole sun]

So it's $L+\epsilon_f$ and $L-\epsilon_f$

[bronze saffron]

ElectricTortoise

[nova heron]

Well, no, for any e_g and e_h, if 0 < |x - a| < min(d_g, d_h) we get that inequality.

And then, yes, given some target e_f we then choose e_h = e_g = e_f and d_f = min(d_g, d_h).

[whole sun]

Mmm alright...

I don't fully understand it, but I think I get the idea

I'll need time to digest this haha

[fringe blaze]

@whole sun

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