help! See equation inside.
\[\lim_{x \rightarrow \infty} (\sin x- \lfloor \sin x \rfloor) \neq 0\]
you want to prove that ?
yes using epsilon and delta.
I have another Idea ! but I don't know if you're able to understand it !
ok. We'll let's try it anyhow. See how it goes.
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 !
you are not familiar with this , Aren't you ?
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.
I didn't understand what are you trying to say !
just to make sure ! write the definition of the limit first ! for sin(x)-[sin(x)]
we will do it together kk @Compgroupmail
ok let's do it then.
first write the definition using your equation editor ( I know it's painful but...it makes things a lot easier ) kk I'm waiting !
which definition? I need to proove it using epsilon and delta.
yeaah ! write this for all epsilon.........
I should say type it !
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 \[\sin x -\lfloor \sin x \rfloor=0\]
but if it is anything else, then it will equal a number \[\neq 0\]
this is not a proof ! :) !
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?
no ! you'll go to jail if you say that is "logical" again !
you didn't want to start ! Just type this Damn definition :) :) !
I honestly do not understand what you are talking about!!!
you don't understand this : \[\forall \epsilon>0.........\]
I do.
so type it ! to make a real proof together !
for \[\epsilon>0 there \exists \delta> \left| x-a \right| that will exist for all \epsilon>\left| f(x)-L \right|\]
this is not the good one ! a =+ infinity f(x)=sin(x)-[sin(x)] L=0 and this what you were supposed to type : \[\forall \epsilon >0 \exists M >0 x \in \mathbb{R} x>M \left| \sin(x)-\lfloor \sin(x) \rfloor \right|<\epsilon\]
are you agree with that ?!
if you didn't understand something ! Just let me know kk :) :) ! @Compgroupmail
still there ! Let's continue ?
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.
I am going to try an easier problem, but thanks anyhow. Iwill rate you best answer! :)
It can't be ! I don't do that for medals...It's just a stupid button !
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!!!
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 !
I need it for today, so later won't be good, but thank you anyways!!!
I'll do it now ! don't worry !
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 !
Join our real-time social learning platform and learn together with your friends!