Question

help! See equation inside.

Answer 1

\[\lim_{x \rightarrow \infty} (\sin x- \lfloor \sin x \rfloor) \neq 0\]

Answer 2

you want to prove that ?

Answer 3

yes
using epsilon and delta.

Answer 4

I have another Idea ! but I don't know if you're able to understand it !

Answer 5

ok. We'll let's try it anyhow. See how it goes.

Answer 6

let's suppose that \[\lim_{x \rightarrow +\infty }\sin(x)-\lfloor \sin(x) \rfloor=0\]

\[x_n=2n \pi + \frac{\pi}{6}\]
well \[\lim_{n \rightarrow +\infty}x_n= +\infty \]
then \[\lim_{n \rightarrow + \infty }\sin(x_n)-\lfloor \sin(x_n) \rfloor=0\]
but \[\sin(x_n)-\lfloor \sin(x_n) \rfloor=\sin(2n \pi+ \frac{\pi}{6})-\lfloor \sin(2n \pi+ \frac{\pi}{6}) \rfloor=\frac{1}{2}-0=1/2\]
then lim 1/2 = 0 that's impossible !

Answer 7

you are not familiar with this , Aren't you ?

Answer 8

not really no. Sorry.
But how about this angle. Is this true?
If sin x is equal to \[\lfloor \sin x \rfloor\] then the limit is zero. But if it is in the range of \[-1\le sinx \] then it can not be true.

Answer 9

I didn't understand what are you trying to say !

Answer 10

just to make sure ! write the definition of the limit first ! for sin(x)-[sin(x)]

Answer 11

we will do it together kk @Compgroupmail

Answer 12

ok let's do it then.

Answer 13

first write the definition using your equation editor ( I know it's painful but...it makes things a lot easier ) kk I'm waiting !

Answer 14

which definition? I need to proove it using epsilon and delta.

Answer 15

yeaah ! write this for all epsilon.........

Answer 16

I should say type it !

Answer 17

I can prove it intuitively, but I need help writing it into a proof. Here it the proof:
If sinx=1 or -1 or 0 then it is true that \[\sin x -\lfloor \sin x \rfloor=0\]

Answer 18

but if it is anything else, then it will equal a number \[\neq 0\]

Answer 19

this is not a proof ! :) !

Answer 20

however, because x approaches infinity, there is no limit on sin x because it will keep on repeating itself to different numbers.
I know this isn't a mathematical proof, but a logical one! Now I need help writing it to a mathematical one!
This question is considered very hard yes?

Answer 21

no ! you'll go to jail if you say that is "logical" again !

Answer 22

you didn't want to start ! Just type this Damn definition :) :) !

Answer 23

I honestly do not understand what you are talking about!!!

Answer 24

you don't understand this : \[\forall \epsilon>0.........\]

Answer 25

I do.

Answer 26

so type it ! to make a real proof together !

Answer 27

for \[\epsilon>0 there \exists \delta> \left| x-a \right| that will exist for all \epsilon>\left| f(x)-L \right|\]

Answer 28

this is not the good one ! a =+ infinity
f(x)=sin(x)-[sin(x)]
L=0 and this what you were supposed to type :
\[\forall \epsilon >0 \exists M >0 x \in \mathbb{R} x>M \left| \sin(x)-\lfloor \sin(x) \rfloor \right|<\epsilon\]

Answer 29

are you agree with that ?!

Answer 30

if you didn't understand something ! Just let me know kk :) :) ! @Compgroupmail

Answer 31

still there ! Let's continue ?

Answer 32

I understand, but I can't do this.This is more advanced than my class and I can't use materials that they didn't have.

Answer 33

I am going to try an easier problem, but thanks anyhow. Iwill rate you best answer! :)

Answer 34

It can't be ! I don't do that for medals...It's just a stupid button !

Answer 35

It's very late for me(been up all night) and I have to get some sleep, but I still have some more problems to finish!!!

Answer 36

ohh ! kk I was just Trying to do it Step by step ! with you ! and this was what you learned a bout lim in infinity I'm sure about that ! If you want ma to do it and you'll check later ! I can do it for you ...Good luck !

Answer 37

I need it for today, so later won't be good, but thank you anyways!!!

Answer 38

I'll do it now ! don't worry !

Answer 39

let's choose \[\epsilon =\frac{1}{7}\]
then \[\exists M>0 \] wich \[\forall x>M \left| \sin(x)-\lfloor \sin(x) \rfloor \right|<\frac{1}{7}\]
let's take \[x=2 \pi (\lfloor M\rfloor+1)+ \frac{\pi}{6}\]
so all we have to do is to see if we have really x>M
\[\lfloor M \rfloor+1>M,,,,,,2 \pi >1\]
then \[2\pi (\left[ M \right]+1)+\frac{\pi}{6}> M + \frac{\pi}{6}> M\]
then \[\left| \sin(2 \pi (\lfloor M \rfloor+1)+\frac{\pi}{6}-\lfloor \sin(2 \pi (\lfloor M \rfloor+1)+\frac{\pi}{6} \rfloor \right|=\left| 1/2-0 \right|<1/7\]
then we will get 1/2<1/7 wich is impossible ! (for epsilon you can choose any number smaller than 1/2) and re-make your own proof !