Proof for lim f(x)g(x) = lim f(x) lim g(x)

[amber frigate]

$[
\text{Let } g(x) \leq M. \quad
\text{If } |x-a| < \delta{1}, \text{ then}
]
[
|f(x)g(x) - Ll|
= |f(x)g(x) - Lg(x) + Lg(x) - Ll|
\leq |f(x)g(x) - Lg(x)| + |Lg(x) - Ll|
]
[
= |g(x)|,|f(x) - L| + |L|,|g(x) - l|
\leq M|f(x) - L| + |L|,|g(x) - l|.
]

[
\text{We may also find } \delta{2} \text{ such that }
|f(x) - L| < \frac{\varepsilon}{2M}
\quad \text{when } 0 < |x-a| < \delta{2}, ; x \in \mathcal{D}{f}.
]

[
\text{If } L = 0, \text{ then we conclude }
\delta = \min{\delta{1}, \delta{2}}.
]

[
\text{Otherwise, if } L \neq 0,
\text{ then by the definition of a limit there exists } \delta{3}
\text{ such that } |g(x) - l| < \frac{\varepsilon}{2|L|}
]
[
\text{whenever } 0 < |x-a| < \delta{3}, ; x \in \mathcal{D}{f}.
]

[
\text{Thus we take } \delta = \min{\delta{1}, \delta{2}, \delta{3}}.
]

[
\text{In both cases we conclude that }
|f(x)g(x) - Ll| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon
\quad \text{when } 0 < |x-a| < \delta, ; x \in \mathcal{D}{f} \cap \mathcal{D}{g}.
]$

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[amber frigate]

damn why isnt my LaTeX working

wait
I'm kinda of new to LaTeX, I dont know why it isnt showing what I want it to show 🥲, its ommiting certain parts of the text

I get the first part of the proof, but my misunderstanding starts after the first whitespace. Why do we take f(x) - L lesser than e/2M? I dont see its relevance in the proof. I also do not understand why we take it lesser than e/2L later, and where does the epsilon/2 suddenly come from?

---

Does anyone know how I can make this LaTeX render properly?

[last kelp]

Well why we can do that is because, we said for every positive epsilon it should hold, so if its divided by 2M, no difference
And for what is the relevence, as you see in the proof, M is multipled by |f(x) - L|
so we can say
$ M|f(x) - L| \le \frac{\epsilon}{2}
As you see in the bottom of the proof,

Wow, latex fail

[amber frigate]

u forgot ur $

[last kelp]

Oh

Well you get the idea

$ M|f(x) - L| \le \frac{\epsilon}{2}

[amber frigate]

it needs to be bound between $

$ hello $

[last kelp]

...

[amber frigate]

oh lol maybe not

[last kelp]

I dont think so

[amber frigate]

$\leq$

[supple sail]

Kobe

[last kelp]

The bot is off

Nvm

$M|f(x) - L| \leq \frac{\epsilon}{2}$

[supple sail]

Will Jefferson

[amber frigate]

I think I understand it a little better now

[last kelp]

Limits are always hard to understand

[amber frigate]

So what we do in most epsilon delta proofs for operations of limits is first defining the individual limits to be something less than e/n, and we choose an n such that when we do algebraic manipulation on them, the addition of these will be equal to e?

e being epsilon here

[last kelp]

Yep

[amber frigate]

Luckily I think this one was the hardest

Or no maybe the g(x)/f(x) is a little harder

Do you mind if I ask a question about that one?

[last kelp]

Sure

Ask :‌)

You can write
h(x) = 1/g(x)

[amber frigate]

For defining this operation we use the equation $\exists \delta \text{for which} |f(x)| > \frac{L}{2}$

dammit

whats the code for there exists

found it

I dont quite understand what this even means

Or is the L here not a limit and are they just fucking me over

wait can you even see this

[last kelp]

?

[supple sail]

Kobe

[last kelp]

What operation?

[amber frigate]

that lim g/f = lim g / lim f

[last kelp]

Just let h(x) = 1/g(x)

[amber frigate]

My textbook is in Dutch, otherwise id have sent you a picture

[last kelp]

then since we proved for f(x)h(x)

Just g(x) musnt be 0

I dont know the proof your book used, lemme search up real quick

[amber frigate]

then it chooses |f(x) - L| < eL²/2

thats basicaly the entire proof

but it starts by saying that 1/f(x) < 2/L

But just that line doesnt make much sense to me

f(x) > L/2 in other words

How can a function be greater than a L/2 at any point? L is not even defined here no?

[last kelp]

You cant really say that

[amber frigate]

They say that the L there is a constant such that f(x) > C > 0, but then they treat it like a limit? (where C = L/2)

[last kelp]

Well I havent seen a proof like this

[amber frigate]

btw they just want to prove lim 1/f(x) = 1/L I checked

[last kelp]

Oh

[amber frigate]

Maybe you dont need to understand the words

Its my first ever rigorous math class haha, i'm sorry if im not describing it well enough

Just a little lost

[last kelp]

They kinda are using L/2 as an epsilon, since for every epsilon positive it should work

I found this proof online

It kinda explains more

Try this proof

I brb

[amber frigate]

I'll look at it in a second, something came up rn

[amber frigate]

Oh wow this one makes so much more sense

Thankyou for finding this for me

are you a mathematician by any chance?

[last kelp]

No, I am just a math lover :‌)

[last kelp]

Your welcome :‌D

[amber frigate]

Did you study any math related subject?

Or are you purely self taught

[last kelp]

Self taught

[amber frigate]

Ah damn okay

Are these like simple proofs in your experience?

[last kelp]

Wbu?

[amber frigate]

I'm first year physics bachelors

[last kelp]

Well simple to understand, no
but these are proofs I did a long time ago

[last kelp]

Nice o.o

[amber frigate]

So youve gotten much further than this?

Glad to know its normal to struggle with this a little atleast

[last kelp]

Further is not really a good word, like mathematics is big, I just like learning something every now and then :‌)

[amber frigate]

Have you done real anylisis for derivatives?

[last kelp]

Yeah

[amber frigate]

Is the diffeculty analogous to limits/continueity?

As in the proper defining and proving of them and their operations, not the calculating limits / derivatives

[last kelp]

I found derivatives simpler than limits and continueity

Integrals a bit harder

[last kelp]

Will also extend to transformations

[amber frigate]

I think starting from integrals we split from the math group so dont define that rigorously, luckily

[last kelp]

You got derivatives though right?

Riemann integral definition is not that difficult

[amber frigate]

Yeah I saw the riemann integral in highschool

[last kelp]

It looks difficult but the concept isnt

[amber frigate]

yes derivatives is the last thing that they will have the rigor I think

[last kelp]

I doubt integrals like darboux or labesgue are taught to math group either

So yeah its both riemann I think

[amber frigate]

Well the prof said the supremum principle would proof very important for defining integrals

[last kelp]

Derivative isnt really hard

[amber frigate]

ah no we still define sequences with some rigor as well

after that its integration

[last kelp]

Oh

Good thing about derivative though, is that its available like almost all the time

Integral isnt

[last kelp]

Not in general for every existing function in the universe, but for most questions I meanr

[amber frigate]

oh yeah for sure

calculating derivatives is at most just lot of computational work

sometimes im just staring at an integral not even knowing where to start though lmao

[last kelp]

Xd

Like the annoying thing is, integrals are almost unpredictable

Like I see a lot of pattern in matrix, in graphs, in topology

Integral I barely see any, idk might be my bad

[amber frigate]

no you're totally right

especialy those complex trig integrals

Where you to apply 5 identities and then the t formulas

[last kelp]

Yeah, those are annoying

[amber frigate]

Anyways thanks a lot for helping me

[last kelp]

Though the most annoying integral I ever saw were these

[amber frigate]

oh damn wtf

I thought the () thing was a sign for combinations?

[last kelp]

Like the first one, can be solved to 2ⁿ, the next one, after spending a lot of time I got
f(n) = 3ⁿf(0)
But then f(0), instead of nicely becoming 1, became -∞😭

[last kelp]

Yep

Its that

[amber frigate]

oh damn didnt even know about taking combinations of more than 2 elements

Does it have any use?

[last kelp]

Well to take y elements out of x elements, then take z elements out of the taken elements

[last kelp]

Though the way I was first introduced to these, is like, they are pascals!

Pascal 2D, Pascal 3D, ...

Call one row x, and a column y

It kinda makes pascal for you

[amber frigate]

ah it expands the pascal triangle into an n dimensional shape?

[last kelp]

Yep

This is 3D pascal, the sum rule holds too

[amber frigate]

thats pretty neat, though I hope I never see it in any class 🤣

[last kelp]

Except triangularly

[amber frigate]

Do you need the gamma function for the first one? or any of these?

[last kelp]

Xd
Its not really a popular topic, it was like some random thing I liked, long ago, I spent the whole summer proving you can creating pascal 4D, by piecing together pascal 3D's in a unique way

[last kelp]

Yep, well it doesnt do anything though, you can also use the reflection identity of euler with factorials to
x!(-x!) = (πx)/(sin(πx))

[amber frigate]

oh damn never seen that identity before

[last kelp]

its actually Gamma(1-x)Gamma(x) = π/(sin(πx))

Originally

But, same as what I wrote

[amber frigate]

I'm gonna save that one, seems interesting

though I wont even begin to try and solve the bigger ones

Where did you find this problem? Do you know for certain they got a solution?

[last kelp]

I gave up after I got negative infinity lol

[last kelp]

I spend time thinking just to make problems lol

[last kelp]

I saw the first one, I was like wow, 2ⁿ, and since these represent pascals and pascals of n dimension all have sum rule thing, you can say that for every single one of them that
f(n) = dⁿ × f(0)
Where d is the dimension, but yeah integrals are unpredictable so I got f(0) = 1 for the first one, giving a nice answer, for the second one I got f(0) = -∞

Though there is another definition for that 3 combination thing

Which is the bottom one, which gets f(0) = 1, but I dont think the sum rule holds

[amber frigate]

Whats the sum rule

[last kelp]

Like in pascal 2D if you some two numbers which are after each other you get the number below

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

[last kelp]

In 3D same holds but the sum rule is not two numbers its 3, in a form of an upside down triangle

in 4D its an upside down pyramid like thing

and so on

Since the first combination thing, gives the values of pascals, sum rule holds for it too

Like this is for the first one

The sum rule is the center one

[amber frigate]

Why is Kn = 2nK0 here?

[last kelp]

In integrals from negative infinity to positive infinity, moving the function by 1 units to left or right wont change the area under it, it can also be written as x+1 = y => dx = dy
well then 2(n,x) = (n+1,x), which is just f(n) = 2f(n-1) = 4f(n-2) = ... = 2ⁿf(0)

[amber frigate]

Hmm

Okay I think i understand that a little

Its a pretty neat problem

Do you ever post ur solutions to these on math platforms?

[last kelp]

I do, though most the time, its for problems, I usually discuss solutions with friends 🙂

[amber frigate]

+close

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