How is this allowed?

How is this allowed?

It's division by 0, how is that finite?

Because sin 0 = 0. As x -> 0, sin x -> 0, and sin x can be approximated by x with small enough x.

This honestly belongs here

Dividing by zero isn't "forbidden", it just usually doesn't give any meaningful answer. In the case of sin 0, however, this has a non-finite outcome, so is meaningful.

More precisely, [math]sin \ x \le x[/math] for all [math]x \ge 0[/math], so (with positive x) [math]\frac{sin \ x}{x} \le frac{x}{x}=1[/math], and so taking the limit as x tends to 0 from the right gives that the limit is finite.

Division by zero is undefined and so is [math]\frac{sin(0)}{0}[/math].

Just because a finite limit exists at a certain value doesn't mean the function is defined at that point.

It's not division by zero, it's indeterminate: 0/0. The limit as x approaches zero is 1 though. Take a look at the graph on wolfram alpha. You're probably in a high school calc 1 course, so when you learn about the derivative and l'hopital's rule you'll be able to evaluate that limit. For your purposes though, "dividing by zero" will yield a one/two sided limit of positive or negative infinity.

>How is this allowed?
It isn't. But this is a removable singularity so you can "fix" it.

(0, inf). The 0 is not included in the interval. Therefore it is allowed

It's the sinc function you uneducated tard

>>What is squeeze theorem for 200, Alex.

>removable singularity

Either you're an idiot or I am. How is it a removable singularity?

u have to divide by my sweet digits

(0,2) means that x = 0 is not included in the set
[0,2) means that x = 0 is included in the set

Not the guy, but consider
[eqn]\frac{\sin x}{x} = \frac{\sum_{k = 0}^\infty \frac{(-1)^k x^{2n+1}}{(2n+1)!}}{x} = \sum_{k = 0}^\infty \frac{(-1)^k x^{2n}}{(2n+1)!}[/eqn]
There are of course some details to figure out (e.g. convergence radius), but that is the idea.

You're the idiot I'm afraid. Expand sin into its series representation and then divide by x, singularity removed.

Alternatively just defining f(0) = 1 also fixes things as it turns out if you can just insert your missing point and still have a holomorphic function on the other end it's a removable singularity

Aw nuts.

sinc is sin(pi*x)/(pi*x) retard

>Division by zero is undefined
It's perfectly-well defined,
it's merely indeterminate
because incomputable.
Lrn2define fgt pls

You don't need calc. All you need is squeezing theorem.

Besides, I don't see how it's relevant; it's an open interval, not a closed one.

It still matters that the limit exists at 0 even if it is an open interval there.

Taylor expand the top and bottom. You're not actually "dividing by zero," you're making closer and closer guesses at what that function "would be" at zero. Because of this, a Taylor expansion argument is perfectly valid.

An easier example is the function

[math]f(x) := \dfrac { x } { x } [/math]

The expression [math] \dfrac { x } { x } [/math] is not defined at x=0.

However, at all values except x=0 the function equals

[math]f(x) := 1 [/math]

and so we say at x=0 there is a removable singularity. If you just define the value f(0) to be 1 as well, the resulting function f is continuous.

The function

[math] g(x) := \dfrac { \sin (x) } { x } = \dfrac { x } { x } + \dfrac { \sin (x) -x } { x } = \dfrac { x } { x } + \dfrac{1}{6} \dfrac { x^3 + \dots } { x } [/math]

is just a more complicated version of that.

L'Hôpital:
[math]
\lim_{x\to 0} \frac{sinx}{x} = \lim_{x\to 0} \frac{cosx}{1} = 1
[/math]

Nah man that's the normalized sinc function the un-normalized sinc function is just sin(x)/x

then what is ]0,2]?

fuck off
this is just the derivative
lim in 0 of (sinx - sin0) / (x - 0) equals sin ' (0) = cos (0) = 1

Division by zero is literally an undefined case of the division operation, which in turn is just multiplication by inverse, as 0 has no multiplicative inverse. 'Indeterminate' is not even a term unless you're working with limits.

Well my friend, the division by 0 is perfectly defined in R as well as in C: it is impossible. This division is defined as impossible. The reason is the one you stated: 0 does not have a multiplicative inverse.

I still dont' know why brainlet english-speaking fags stil say "undefined" as it is just "impossible". It. Is. Not. Possible. This is quite a defined statement, isn't it?

Please stop. Neither 'Impossible' nor 'possible' are mathematical words.

Oh oh, we apparently have the visit here of a specialist of the semantics of mathematics? Nice. I should have said "the inverse of 0 does not exist". You can't disagree with that then.

If you define the notion of "possible" by the possibility of existence, then the inverse of 0 is not possible, because it does and will not exist. It is therefore "im-possible", as the shorter way of saying "not possible".

Go drink your tea and go to bed now, son.

No, it is possible to define division by 0. You can define it as a/0 = 1 for every a, for example. Defining division for everything that has a multiplicative inverse and defining division by 0 separately is not "impossible". Its just a definition; we can define things however we damn well please. What you cannot do is define division by 0 in any way that respects the other field properties of R or C. I hope your little pedant brain can understand all of that.

Also, when you're working with other objects, division by 0 is very possible if you define it the correct way (for example, by using limits or analytic continuation or divisors). It's not only possible, but it's important to define division by 0 in these cases.

My little brain tells me something. The sets R and C are constructed out of axioms and definitions.

In this environment of definitions and logical statements, with the addition of definitions for "group", "ring", and "field" (and many other notions) these two sets are seen as fields (on R by instance for R, and on R or C for C, for classically).
In this context - which was the context of the discussion, little faggot, your arguments do not hold.
This context is quite general. In any field of any fucking objects (matrices, functions, whatever you want), 0 has no inverse. Division by zero is not possible.

Your argument then is: well, but I can define it as I want.

Ah! Who cares that you can define whatever equals (or not equals, or whatever binary relation you want to work with - you can even redefine what a relation is) whatever. You can redefine the additive law of R, say 0^-1 exists, you can even redefine ZFC. Of course you can. You can define f(x) = sinx/x everywhere but in 0 and sinx/x = your mom's ass.
The question is: will this definition produce anything substantial, coherent?

Fuck off now with your pretentious stupid, impolite and brainless statements. Go to bed I said.

Now, as for other objects and division by 0, if you would like to explain, please do explain. I'm interested and not knowledgable about.

Parentheses notation is fucking stupid, just use brackets.
That would be ]0, infinity[.

Its a removable singularity, so dividing by 0 is okay ( since you're not actually dividing by 0, think of it like x/x). On the other hand 1/x has an essential singularity at 0.

checked

"No"

What are the characteristics of a state function ?

Nigga if you're talking about the set of Q, Z, R, or C, which is where this problem dwells, the sets are closed and you can't divide by 0