Recent content by jessjolt2

ohhh gotcha thanks...are you sure though that this 1 can't possibly follow from some earlier step?? i just want to make sure because it is bothering me lol

haha its legal to just do that? i mean this is supposed to be a rigid proof...i don't think u can just add a random 1 without explaining why? why not just say that m\neq0, l\neq0 ? i think the 1 comes from a more logical step?

But delta#2 is not \frac{1}{2|m|+1} it is \frac{1}{2(|l|+1)}

I know that you have to combine the two deltas by taking the minimum delta, but my question is, how does Spivak get that \left|f(x)-l\right| < min (1, E/( 2(|m|+1) ) ) ? He doesn't get \left|f(x)-l\right| < min (1, E/( 2(|m|) ) He is adding some random 1 next to the |m| ?

Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i don't understand why he is doing something.. ok so the following i understand: \left|f(x)g(x)-lm\right|< E...

Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i don't understand why he is doing something.. ok so the following i understand: \left|f(x)g(x)-lm\right|< E...

why would it not be finite if f(x) did not go to 0? specifically i want to know if my proof is right? my professor wants it to be proved using epsilon-delta, not just words..

Homework Statement It is given that the limit as x->a of ( f(x)/(x-a) ) is 3. Prove using the espilon-delta theorem of limit that the limit as x->a of f(x) is 0. The Attempt at a Solution so it is known that: |( f(x)/(x-a) ) - 3| < E1 when |x-a| < d1 therefore: |( f(x)/(x-a)...

i kind of see your point, but i do not see how this relates to my question? and based on current mathematics, this function is infinitely divisible on the interval (0,1), (2,3), etc, but it is not divisible at all on (1,2), etc... my question is, what if space is not infinitely divisible on...

no? doesnt infinitely divisible mean continuous?

Can't Zeno's paradox be taken as proof that space isn't continuous?

Sorry, I didnt mean that calculus doesn't give good approximations nd that we should stop using it. But i still think its a problem that calculus can't describe nature at small scales if calculus assumes spacetime is continuous if it possibly isn't...my point is that there should be developed a...

Well from my experience it seems like calculus is the main mathematical tool in describing physics... I read Majid's essay (thanks Studiot), and he describes some math that he and others developed to include the discrete nature of spacetime, although its too complex for me to understand lol...

But physics uses math and deals with real space...so it should care if space is continuous or not, at least when it is used to describe motion in space and watnot... Does anyone know of any literature that discusses the relationship between the possible discreteness of space and mathematics?

Ok so mathematically you can divide any number by any other (nonzero) number and you can keep dividing that number however many times you want. Like dividing 1 by 2 and then by 2 again etc. And this is the basis of the famous paradox that mathematically, you can't really move from point a to b...