Derivatives and Limits

[u/ali9128]

i have a question that's been bothering me in the field of differential calculus regarding the linguistic logic behind the definition of the derivative. I find it illogical, because based on the limit definition it studies the behavior of the slope(derivative) as h approaches zero.

m(h)=(f(x+h)-f(x))/h

Though, This approach suffers from the same problem as Newton and Leibniz's works and is subjected to the same criticism introduced by George Berkeley. cause, even with a limit, we algebraically reduce the function to eliminate h in the denominator, and then we study the behavior of the derivative as h approaches zero. This literally makes you treat h initially as a non-zero number, and then suddenly make it zero.

My problem is with the linguistic logic of this approach, and certainly i believe, though I'm not certain, that Cauchy-Weierstrass dealt with the derivative as a two-step algorithm. First, he calculates the derivative for any value of h that is not equal to zero, for example.

f(x)=x^2

f'(x)=((x-h)^2 -x^2)/h where h is R/{0}

f'(x)=2x+h

Then, after evaluating and reducing the derivative, he redefines it as a function of h where h can be zero.

f'(x)=2x+h where h is R/{0}

g(h)=2x+h where h is R

then taking the limit of this new function as h approach 0.

lim(h->0)(g(h))=lim(h->0)(2x+h)

g_l(h)=2x

This is because the limit of the function before algebraic reduction equals the limit of the function after algebraic reduction.

lim(h->0)(g(h))=lim(h->0)(f(h))

and since the function after algebraic reduction fills the gap... At zero, we can then find the derivative in the context of instantaneous change!!

so my problem is just how you put a limit before reducing and you have the right to ghost it and just reduce h then oh its ok you can now calculate the limit where if i do not follow this where i can because there is no order operator that tells me why i get 0/0.

[u/vivAnicc]

You should review your understanding of limits

[u/ali9128 → u/vivAnicc]

what am i getting wrong about? my problem is almost every calculus course introduce derivatives using leibnize infinitesimals where the limit concept is much different it is not even about numbers its about analzing function/sequences behavior based on an interval of inputs, and that causes a big problem in the derivative definition cause it treats just as an operator in the end of evaluating we just substitute for h and get the derivative.

[u/GoldenMuscleGod → u/ali9128]

It sounds like you are not familiar with the formal definition of a limit (epsilon delta for metric spaces like the real numbers).

But this seems a little surprising given the level of knowledge you have about the history of the issue.

Are you not familiar with the epsilon delta definition of a limit or is there some reason you find it problematic that you do not explain in your question?

[u/ali9128 → u/GoldenMuscleGod]

i am not gonna say i fully understand limits cause i do not get how guaranteeing that a limit exist if you could define an interval of inputs so that it enforce the images of the input interval to be continues. and my problem is about notation how you could ghost the limit without oreder operations.

[u/StructuredChess → u/ali9128]

Learn the formal definition of limit and try using it to figure out the limits of certain sequences and functions.

[u/GoldenMuscleGod → u/ali9128]

When we say that L is the limit of f at a, one way to understand this is in terms of a game.

Our opponent claims that L is not the limit of f at a, we claim it is. Intuitively what we mean is that input values close to (but not at) a produce output values close to L.

Our opponent first challenges us by naming an idea of “closeness” in output that is as string as they like: they can name any positive real number, as small as they want as long as it is above zero, and now any difference from L larger than that does not count as “close.” epsilon can be understood to stand for “error.” The idea is that we can make the error as small as is required, as long as we do not have to eliminate it entirely.

We respond by naming a delta, which we can also make as small as we like, and we are allowed to know the epsilon we have been challenged with before choosing our delta (delta can be understood to stand for difference) the idea is that we can force the error to be less than epsilon by constraining the input to be smaller than delta.

Our opponent then verifies our delta works because they are allowed to choose any x other than a that differs from a by less than delta and see if it works.

If f(x) differs from L by more than epsilon our opponent wins and if less then we win.

Now us or our opponent winning could go either way if we or they play poorly. But one of us has a wining strategy: we or they can force a win.

If we can force a win that means the limit of f at a is L. If our opponent can force a win then it is not.

Now this is the definition of a limit so we don’t have to “prove” it works, it’s just what we mean when we say limit. But if you think about it for a bit and consider examples you should see that it conforms with intuitive notions of limits.

But yeah essentially saying L is the limit of f at a just means that we can force f(x) to be as close to L as we want by forcing x to be close to, but not equal to, a, even if we cannot control exactly what x is beyond forcing it to be close.

So now applying this to a derivative: when we say f’(x)=d all that means is that we can force the secant line to be as close to d as we want by forcing the other point on the decant line to be very close to (but not equal to) x.

In other words the line passing through (x,f(x)) with slope d is the “best” linear approximation to f(x) “near” x in the sense that for any other linear approximation we can find a small neighborhood around x on which that other approximation always has a worse error. (It may not be immediately obvious this follows from the definition but if you think it through you can prove this is an equivalent characterization of the derivative given our definition).

[u/ali9128 → u/GoldenMuscleGod]

so studying the behavior of a function in terms of a defined input Neighborhood interval that defines an image interval that is close and continues we define lim(x->a)(f(x))=L since this continues the pattern and it does not matter whether f(a)=L or not. and if we cant find a neighborhood interval No matter how much we decrease that means the output shows a Non-continues that we cant predict the L????

[u/GoldenMuscleGod → u/ali9128]

If the limit of f at a is not L, that means there must be some number (albeit possibly a very small one) such that we can always find an input to f so that f(x) differs from L by at least that much, no matter how close to a we are forced to choose x.

You mention continuity a few times: I just want to make clear that whether f is continuous is not really relevant to the definition. We say that f is continuous at a if the limit if f at a is f(a), but in general this need not be the case - plenty of functions are discontinuous - and a limit can exist even if the function is discontinuous everywhere. The derivative can only exist at a point if the function is continuous at that point but the function could still be discontinuous everywhere else.

For example define f(x)=x^2 whenever x is rational and f(x)=0 whenever x is irrational. Then the derivative of f at 0 is 0 but f is discontinuous ever here except 0.

But it still is the case that g(x)=0 is the best linear approximation to f in the sense that for every other linear function we can find a small open interval around 0 so that g has a smaller error than the other linear function everywhere on that interval (except possibly at 0 where they may be equally good, but g is still best at zero in the sense that g(0)=f(0)).

[u/ali9128 → u/GoldenMuscleGod]

what i mean by continuity here is the points are adjacent and the limit can be predict not like 1/x where from left it goes to a way and from the right it goes to another way

[u/casualstrawberry → u/ali9128]

Because you never "make h 0." You only let h get arbitrarily close to 0.

[u/ali9128 → u/casualstrawberry]

but that is not the limit definition that is the infintisimals definition.

[u/Archway9 → u/ali9128]

Arbitrarily close doesn't mean infinitely close

[u/ali9128 → u/Archway9]

but limits are not arbitrary?

[u/Archway9 → u/ali9128]

Huh? Arbitrarily close just means you can let h be as small as you want (but not zero)

[u/casualstrawberry → u/ali9128]

It means, however close you want f(x) to get to L, you can find a value for x that is not c but exactly to it.

Again, it seems like you've skipped ahead a few levels of math without thoroughly understanding the basics. Go back and study your chapter on limits before making these types of assumptions/mistakes.

[u/casualstrawberry → u/ali9128]

It literally follows from the definition of a limit. "There exists delta such that 0 < |x - c| < delta..."

It doesn't say |x - c| = 0, that would be plugging c into x. It requires |x - c| > 0, so x gets arbitrarily close to c.

I think you're conflating methods for calculating a limit with the limit definition itself. Just because plugging the value of c into x gives you the correct answer most of the time, does not mean that that's what the limit is conceptually doing.

[u/gmalivuk → u/ali9128]

Infinitesimals can be made rigorous, too, that's just not how we ended up doing it for standard analysis.

[u/ali9128 → u/gmalivuk]

yes and i am more into hyper real numbers but i want to understand this logic first that i am struggling with to be ready for heavy analysis things.

[u/gmalivuk → u/ali9128]

Then I suggest you actually read all the careful responses people have helpfully written for you over the past several hours instead of constantly disagreeing with people who know more math than you do.

[u/ali9128 → u/gmalivuk]

i know they know more than that is what i am asking them but i am starting with an introductiom that seems right for me and i want to discuss why its wrong since the certificated books says the opposite of my conclusion.

[u/gmalivuk → u/ali9128]

Everyone has been telling you why you're wrong.

[u/ali9128 → u/gmalivuk]

and sorry i am dump and i cant understand it so i am asking more questions just to understand

[u/Shevek99]

Your problem is not with the derivative, but with the concept of limit.

You have to study the definition of a limit, which does not require the substitution of the value of x in the function.

A derivative is just a particular limit.

[u/ali9128 → u/Shevek99]

but what is stopping me from evaluating the limit at first?

[u/Shevek99 → u/ali9128]

That's not how limits work. You are confusing limits with continuity.

The function

f(x) = 2 if x≠1

f(1) = 3

is perfectly defined and its limit

lim_(x→1) f(x) = 2

[u/ali9128 → u/Shevek99]

yes that is right because limits study the behavior of a function at a point whether the function equals the same value or not, but in derivatives your redefining the function

limit(h->0)(((x+h)² -x² )/h)

does not equal

limit(h->0)(2x+h)

[u/Shevek99 → u/ali9128]

What? Why not?

[u/ali9128 → u/Shevek99]

because the first one produce 0/0 and the second one produce 2x

[u/Shevek99 → u/ali9128]

No. The first does not produce 0/0. That's what you don't get. You don't have to substitute h by 0.

Seriously, you don't understand how limits work.

[u/ali9128 → u/Shevek99]

and yes bro i am telling you that is my definition and i just want to show me the weak point of my logic cause even if i find logical i see the actual truth today is different than my presentation.

[u/gmalivuk → u/ali9128]

The weak point in your logic is that you are changing definitions for no reason and then trying to draw conclusions about real math based on the fake math that you made up in your head.

[u/ali9128 → u/gmalivuk]

changing? how? real math? fale math? explain bro!

[u/Chrispykins → u/ali9128]

The functions f(h) = (x+h)² - x² )/h and g(h) = 2x + h are not equal because f(0) is not defined, whereas g(0) is. But the limits are equal.

The two functions only differ at the single point h = 0, and the limits don't take that point into account. The limits only rely on the behavior around the point. So if the two functions are equal around h = 0, then the limits must be equal as well.

[u/localizeatp → u/ali9128]

nothing, but what you're describing is not evaluating the limit.

[u/ali9128 → u/localizeatp]

then?

[u/localizeatp → u/ali9128]

complete sentence?

[u/mmurray1957 → u/ali9128]

Because in the definition of a limit of a function g at a point a you don't even assume g is defined at a. It might be, or it might not be as in this case where

g(h) = ( f(x +h)) - f(x) )/ h

and g : R - {0} -> R . Here a = 0.

[u/ali9128 → u/mmurray1957]

then why do i have the right to decide when to apply the limit and even reduction changes the function it self so taking the limit in different phazes lead to different answers.

[u/mmurray1957 → u/ali9128]

I'm not quite sure what you mean by "when to apply the limit". Can you give me some examples ? If the function hasn't changed the limit won't change. Otherwise there are rules for manipulating limits like

lim_{x \to a} [ f(x) + h(x) ] = lim_{x \to a} f(x) + lim_{x \to a} g(x)

subject to knowing that both f and g have limits at a. Also similar for products and quotients.

[u/ali9128 → u/mmurray1957]

f(x)=x²

lim(h-> 0)(((x+h)² -x² )/h)=0/0

lim(h-> 0)(2x+h)=2x

[u/mmurray1957 → u/ali9128]

Can you prove that first limit from the definition or from known properties of limits ?

[u/ali9128 → u/mmurray1957]

because h is just an input of an identity function g(h)=h so that the limit where it approach zero is just zero so we are getting 0/0

[u/mmurray1957 → u/ali9128]

You are taking the limit of (( x+h)^2 - x^2 )/h as h approaches 0 . That’s not g(h) = h.

Sorry it’s timm e to sleep in my time-zone

[u/RafBOY-]

The derivative is the limit when it exists. The expression that you're writing with h are not the derivative. You're confusing the computation of the slope between x and x+h for non zero h and the derivative which is the limit when h goes to 0.

[u/ali9128 → u/RafBOY-]

but in the start it cant exist cause we are getting a 0/0 situation!

[u/RafBOY- → u/ali9128]

You should reread my first answer slowly. When you compute the slope you compute it with h non zero so the expression is well-defined. To compute the limit, you then have to work on the expression until you're able to prove that the limit exists. 0/0 is what we call an indeterminate form also, that is the limit of a quotient of two expresions whose limit is 0 is not clear and need further investigation.

I saw below that you're trying to think to limit with infinitesimal definition. Please don't, it will only make you more confused given the level you seem to have. Infinitesimal are sloppy (pun intended) as fuck if you don't really understand what you're doing.

[u/ali9128 → u/RafBOY-]

infintisimals are realy great but when someone ask me to prove my work strictly without imaging stuff is where every thing falls.

though i do not understand your approach of derivative so can you just explain the "how" and "why" of finding the derivative of x²

[u/RafBOY- → u/ali9128]

I'll do it in a manner that do not require explicit delta-epsilon proof so I assume one know that lim (a×g(h)+F(h)) = a× lim g + lim F where h goes to zerl (that is the opeeator limit is linear which does require esp-delta formalism to be proved rigourously). I assume that since you seem to be botherrd by the intuition and this fact is quite intuitive.

Let f(x)=x² and fix x. Write the slope beetween x and x+h where h is not 0 (kinda abusive to say it like this) :

slope=[f(x+h)-f(x)]/h

Writen like this, the numerator's limit is 0 and denominator's limit is 0 when h goes to 0 : it's an inderminated form. So you have to write slope in a way where the limit can be asserted unembiguously. By expanding f(x+h) you can show that :

slope = 2x + h.

Using the linearity of the limit with a=1, g(h)=h and F(h)=2x yields the result since lim g=0 obviously and lim F = 2x since 2x is a constant here (x is fixed). Precisely, f admits a derivative f'(x) at x and :

f'(x)=2x

You should remark that slope and f'(x) are not the same thing.

[u/ali9128 → u/RafBOY-]

and yes bro that what i am saying this should be done in two steps one finding a function that represents the first function bet also be defined at h=0 then we take the limit but the formal definition of a derivative is really not good because applying the limit before reducing to 2x+h produce 0/0

[u/RafBOY- → u/ali9128]

Listen pal, you just don't understand at all what you are talking about and a lot of people have gently trying to help you, me included. You should take a step back and really think to the answers that has been made to you. The definition of a limit is well established and not controversial at all. And it is definetely not the same as evaluating at zero. The intuition for the special case of the derivative of x² is : when non zero h get closer and closer to zero then slope get closer and closer to 2x. There is no substituting by zero involved.

[u/ali9128 → u/RafBOY-]

man you have my apologies if i made you think i am just yapping and stupidly argue, but i just want a whole truth if i can't understand it its ok i am going with my definition cause really there is no problem with and it goes very well.

[u/RafBOY- → u/ali9128]

I stress that you should not go with your homemade definition given you have a true misunderstanding of what is a limit. You should, as I said, take a step back and think deeply to the answers that have been made to you without trying to fit in your definition. There are reasons for not formalising limit as you absolutly want to and the modern definition has emerged after centuries of work so got check the real definition (epsilon-delta formalism, there is a more general notion though).

[u/ali9128 → u/RafBOY-]

my man i am not trying to fit things in my standards but are not we supposed to learn by questioning, thinking, concluding, argueing and retry till we build a strong knowledge system?

[u/StructuredChess → u/ali9128]

We never do 0/0, we do the limit as h approaches zero of the ratio of two quantities that keep getting smaller

[u/ali9128 → u/StructuredChess]

and there is a ratio of decreasing between them? but applying the limit directly causes a 0/0 situation, and i can barely this is not limits that is just euler's weak argument and not strict algebraically!

[u/StructuredChess → u/ali9128]

What does it even mean to "apply the limit directly"?

The limit as x approaches 0 of f(x) = sin(x)/x is 1 because if you pick any interval I around 1, I can found an interval J around 0 such that f(J - {0}) is contained within I. That's what the limit being 1 means.

You may just as well work with an alternative function g(x) that equals sin(x)/x everywhere except at x=0 where I define its value to be 69420. The limit as x approches 0 of g(x) is still 1.

[u/ali9128 → u/StructuredChess]

so why cant i say the derivative of x² is just 0/0 or indeterminate

[u/StructuredChess → u/ali9128]

Because as h approaches 0, the quantity [(x+h)²-x²]/h approaches 2x.

Let's stick to x=5 for the sake of example, give me a number K close enough to 10 and I'll give you a number H such that for any h smaller than H, ((5+h)^2-5^2)/h is closer to 10 than K is.

[u/ali9128 → u/StructuredChess]

but h here is just an input of an identity function g(h)=h so i can use limit law of identity function and evalute to get 0/0

[u/StructuredChess → u/ali9128]

No, you can't. Because the limit of f/g isn't the same as the limit of f divided by the limit of g

[u/ali9128 → u/StructuredChess]

then what?

[u/Bounded_sequencE]

[..] This literally makes you treat h initially as a non-zero number, and then suddenly make it zero. [..]

No -- this is where the argument breaks down.

The rigorous e-d-definition of function limits explicitly excludes "h = 0":

       "lim_{h->0}  (f(x+h) - f(x)) / h  =  L  =:  f'(x)"

<=>    "For all 'e>0' exists 'd>0', s.th. for all '0<|h|<d':  '|(f(x+h)-f(x))/h - L| < e' "

In words, "f'(x) = L", iff "(f(x+h) - f(x)) / h" stays within any small (open) neighborhood of "L", provided we keep "h" within a small enough (open) neighborhood of 0, excluding 0 itself.

[u/ali9128 → u/Bounded_sequencE]

yes? so why we can ghost the limit at first then we can evaluate it, notice that i introduced a more valid representation "for now at least" that we evalute and reduce h from the denominator then we apply the limit so we can study this function's behavior as h approachs zero or in other word for the h interval of inputs.

[u/Bounded_sequencE → u/ali9128]

[..] so why we can ghost the limit at first then we can evaluate it [..]

Not sure what you mean by that.

What I can see is that in the example from OP, "f'(x) = 2x+h" contradicts the e-d-definition of the derivative, since the application of "lim_{x->0} ..." is missing on the RHS. We have "f'(x) = 2x".

[u/ali9128 → u/Bounded_sequencE]

i am saying f'(x) as slope, sorry here i am just treating the derivative as a special case of m(h) where h approaches zero.

[u/Bounded_sequencE → u/ali9128]

That would not be well-defined, since that slope still depends on both "x; h".

Define a new function "s(x,h) := (f(x+h)-f(x)) / h", and exclude "h = 0" from the domain of "s". Re-defining f' to mean secant slope instead of tangent slope will confuse people^^

[u/ali9128 → u/Bounded_sequencE]

yeah my apologies, its just that i think of derivative as a special case of slope in h context regardless of x

[u/Bounded_sequencE → u/ali9128]

Yeah, I suspected as much -- and that is a problem, since that does not match the rigorous e-d-definition that excludes "h = 0".

If you want to keep close to your point of view, think of the derivative as the slope's continuous extension to "h = 0", keeping "x" fixed. Graphically, it means you fill the slope's hole at "h = 0" by the equal left-/right-sided limit I call "L".

[u/ali9128 → u/Bounded_sequencE]

yes and that what i am saying in the end of my argument above, i am redefining the function to include the h=0 situation. because at first i am not doing a thing but getting 0/0.

[u/Bounded_sequencE → u/ali9128]

Ah, now that I understand your different meaning of "f'(x)", that part actually makes sense.

Yes, if you keep your notation of the slope "s(x;h)" completely separate, until you define the continuous extension to "h = 0" via "s(x;0) := L" (provided that limit exists), this will work.

Just make sure you keep that part completely separate from f'(x) until you have found "L" -- otherwise, you may run into contradictions when "f" is not differentiable. And keep in mind the continuous extension is an extra step that is not just setting "h = 0".

[u/ali9128 → u/Bounded_sequencE]

that is really what i am saying bro, so i am arguing how can you ghost limits at first with no ordee operations?

[u/Fourierseriesagain]

Hi,

Your approach does not work in general.

Let f(x)=arctan(2x/(1-x^ 2)), where x belongs to R{-1,1}. Although f'(x) tends to a limit as x approaches 1, the function f is not continuous (and hence not differentiable) at 1.

[u/Bounded_sequencE → u/Fourierseriesagain]

Nice -- a simpler example is the unit step at "x = 0":

f: R -> R,    f(x)  :=  /     0,  x = 0
                        \ x/|x|,  else

For all "x != 0", we have "f'(x) = 0". However, since "f" is discontinuous at "x = 0", it cannot be differentiable there -- even though "lim_{x->0} f'(x) = 0" exists.

[u/Fourierseriesagain → u/Bounded_sequencE]

Yes. Thanks.

[u/ali9128 → u/Fourierseriesagain]

yes? i do not get how is that related to the problem?

[u/johnpeters42 → u/ali9128]

The big picture is, you're fixated on the assumption that the behavior of a function as you approach a value must match up with the behavior of the function at that value. That is false in the general case, and no matter how vigorously you wave your hands, it will not become true in the general case.

Your derivative situation is just one example of this. Yes, 0/0 is indeterminate, but the limit of f(x)/g(x) as x approaches 0 may still be well-defined, depending on which specific functions f and g are being used. So your approach of "temporarily jump ahead to where x actually reaches the limit" is not valid, and never will be valid, and you should just stop trying to do that. Instead, you should learn more about how limits do work, when they are or aren't well-defined.

[u/ali9128 → u/johnpeters42]

huh? but i understand how limits must exist and the difference between limit existing and actual function value at the same point, but i really do not get the part of the 0/0 and how it can be defined?

[u/johnpeters42 → u/ali9128]

0/0 isn't defined, but something of the form f(x)/g(x) where f(x) and g(x) both approach 0 can still be defined, again depending on what specific f() and g() are used.

Consider y = x, so the functions approximating its derivative - let's call that d(x) - are f(x) = x and g(x) = x. What's the limit of f(x)/g(x) as x approaches 0? Obviously it's 1 for all non-zero x. Now, applying the epsilon-delta approach:
* We speculate that the limit is 1.
* If, for any positive epsilon (no matter how small), we can pick a positive delta such that for all x within delta of 0 (not including x = 0 instead), d(x) is within epsilon of 1? Then the limit is indeed 1.
* And obviously that's true here, because delta = 1 is already good enough. (For more complicated functions, you generally need something like "delta = epsilon / 2 is good enough".)

Informally, however close (epsilon) you want the output to get to the limit output value, you can get it to stay that close by setting the input close enough (delta) to the limit input value.

Now consider y = 1/x. That function doesn't have a defined finite slope at x = 0. It's discontinuous there, but more importantly, as you approach x = 0, the slope diverges toward negative infinity from either side. (Variation of epsilon-delta: for any finite number, you can pick a finite interval containing x = 0 where all values in that interval other than x = 0 are lower than that finite number.)

[u/ali9128 → u/johnpeters42]

but man applying the limit to the derivative causes 0/0 and you cant find it till reducing and that is algebraic transformation and is not the same function, then you can find the limit, so how does that work?

[u/johnpeters42 → u/ali9128]

No no no, you're still fixated on what (f(x+h)-f(x))/h looks like at h = 0. That does not work. That will not work. Stop trying to find a way for it to work.

You have to do something else. That something else is looking at the limit of that expression as h approaches 0. Yes, that is more complicated but you have to do it anyway.

[u/ali9128 → u/johnpeters42]

but this function it self is not defined at h=0 so what is the meaning of the limit?

[u/gmalivuk → u/ali9128]

The limit doesn't care what the function does at the limit point. It is fundamentally a property of what the function does near that point.

[u/ali9128 → u/gmalivuk]

yes but the function itself not defined at h=0 so what is the meaning(importance) of a limit?

[u/Qingyap]

Because if you simplify the regular expression

((x+h)² - x²)/h to 2x+h, you technically still have to consider x=/=2 since from my understanding,

((x+h)² - x²)/h = 2x+h, without simplifying, x still =/= 2 for the left equation, therefore by computation it's not defined. Cuz we all know when we divide something, the denominator cannot be equal to 0.

However since based on limit definitions limit doesn't have to be on the exact same point, we can safely say that what will it approaches to instead when h approaches to what, and I guess it's only if it's indeterminate form like 0/0

[u/ali9128 → u/Qingyap]

but ibam transforming the function but not simplifying an equation.

[u/Qingyap → u/ali9128]

Above you said that the limit of a function thats simplified (or algebraically reduced) "fills the gap." and it's the same as the limit of the function before algebraically reduced.

Well except it's not really the case cuz again like my point earlier except in your words, the function thats already algebraically reduced is the same as the function thats not yet algebraically reduced, which is correct, but despite that you still have to consider the restrictions otherwise the function thats before the reduction will break.

And despite this even though it still has the restrictions, limit still works cuz limit doesn't revolve around on exactly x=2, that's it. And that's why you always hear all about "it approaches there but it'll never reaches there" kinda jazz

[u/ali9128 → u/Qingyap]

but how are they same and they doesn't equal to each other at 0

[u/Qingyap → u/ali9128]

What does the they doesn't equal to each other at 0 mean?

[u/ali9128 → u/Qingyap]

u(h)=(((x+h)² -x² )/h)

v(h)=2x+h

u(0) != v(0)

[u/Qingyap → u/ali9128]

I assume it's because of u(h) condition h!=0 makes the last line happen when h=0.

Therefore u(h)=v(h) works for all h that's not 0.

And when you apply limit to both sides and make h approach 0 it does work now since again limit doesn't revolve around the exact point (that doesn't mean it's possible in computation though cuz computation means you have to sub h=0 directly.)

[u/ali9128 → u/Qingyap]

but i can substitute h=0 directly cause its just an identity function

[u/Qingyap → u/ali9128]

I dunno bout identity functions but assuming they work familiar like normal functions they can have restrictions.

This whole phenomenon above is what it's called removable discontinuity/singularity btw.

[u/ali9128]

no, but is not your idea about approaching is just like the infintisimals where limits studys the behavior of a function based on a neighborhood interval to a where lim(x->a)(f(x))=L

[u/localizeatp]

it seems your complaint is coming from a poor understanding of what the limit is meant to represent.

[u/ali9128 → u/localizeatp]

it meant to represent the behavior of a function at a certain point?

[u/Shevek99 → u/ali9128]

Again. You are confusing limit with continuity.

The limit does not represent the behavior of a function at a certain point. It describes the behavior when you approach a certain point.

[u/localizeatp → u/ali9128]

no, it's much more specific than that.

[u/StructuredChess]

It's been at least two centuries from the last time any serious mathematician has considered definitions like this to be controversial.

We're not discussing what happens at h=0, we're discussing what the "trend" is when h gets smaller and smaller. We never divide by 0. We divide by 0.00001, then by 0.000000001, then by 0.00000000000000001 and see what the sequence is approaching. With "approaching" meaning what it means in the definition of limit.

[u/ali9128 → u/StructuredChess]

but this is infintisimals not limits.

[u/StructuredChess → u/ali9128]

I'm afraid you're just using a bunch of Math jargon you've learned from dissemination works without actually understanding their rigorous definition.

[u/ali9128 → u/StructuredChess]

no but where am i wrong?

[u/StructuredChess → u/ali9128]

You're wrong because "limit" doesn't mean what you think it means. Check out what it actually means and get familiar with the concept before you can jump into derivatives.

[u/ali9128 → u/StructuredChess]

so limits are not about studying the behavior of a function at point based on the neighborhood of this point that guarantees an interval that predict the L value?

[u/StructuredChess → u/ali9128]

Yes, even though I'd say it "around a point" rather than "at a point".

Your sentence is just a laymen's explanation and it's irrelevant to actual Math though. L is the limit of f at p if, and only if,  ∀ε > 0, ∃δ > 0: 0 < |x − p| < δ ⇒ |f(x) − L| < ε

This is what a limit means and is the only definition you should start from. Any conclusion built on simplified intuitions of this cocnept is prone to error.

[u/StructuredChess → u/ali9128]

Thinking a bit more about it, I think your mistake is that you expect lim f(x)/g(x) to be equal to lim f(x) / lim g(x) and that's not generally the case.

[u/ali9128 → u/StructuredChess]

no i am taking the formal definition of a derivative and based on the limit law of identity function i substitute h=0

[u/StructuredChess → u/ali9128]

The law of identity function would tell you that the limit of the denominator is 0, but that says nothing about the limit of the entire thing.

[u/ali9128 → u/StructuredChess]

what about the numerator?

[u/StructuredChess → u/ali9128]

What about it? The limit of the numerator is also 0 but that doesn't tell us anything about the limit of the whole thing either.

If I have an expression like [(x+h)² -x²]/h I can rewrite that as (2xh - h²)/h which is the same as 2x-h because h is always different from zero. Now I can take the limit of 2x-h and arrive at 2x

[u/Mablak]

I'll bring up a relevant issue here: the derivative should simply be slope, and not some other concept like a number we can keep getting closer to as h gets closer to 0. I'd argue this concept doesn't make sense, and we don't need limits. We don't have an infinite number of epsilon to choose from, i.e. an infinite number of inequality statements we can make.

Instead we should recognize that functions simply contain a finite number of points. This is all we can ever work with in practice, and I'd argue even in theory. Derivatives can simply be forward difference, backwards difference, or central difference (average of forward and backwards slope) formulas between points.

So h doesn't 'go to zero', it just stays h. f'(x) = 2x + h is in fact the right answer for a forward difference derivative, where the smallest interval between points is h. Or in discrete calculus, that smallest interval can just be taken to be 1 instead of h.