Proving Limit False I'm trying to prove that the limit of sin x as x->infinity is not equal to 1/2. I know that this is true, but I can't seen to figure out how to prove it using the precise definition of a limit.
What I have so far is this,
e>0, M>0
abs(sin x - 1/2)<e whenever x>M

I also think that I can use the fact that,
abs(sin x)<=1 for all x

But I don't know for sure. I've also seen some really extensive proofs for stuff like sin x/x (ref 1). So I may be doing this completely wrong.
Thanks for any help.


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*http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx
 A: Choose $\epsilon_0 = 1/4$. Then for any $M>0$, choose an integer $K$ such that $K\pi>M$. Then $$|\sin(K\pi)-1/2|=1/2>1/4=\epsilon_0.$$ Thus by the definition of limit, $$\displaystyle\lim_{x\to\infty}\sin x\neq 1/2.$$
A: (Do see comments bellow.)
Well, for this to be true, it must hold that
$$\forall \epsilon > 0 \exists \delta > 0 \forall x \in P^-(\infty, \delta): |\sin x| < \epsilon.$$
Obviously, this does not hold for $x \in \{\frac\pi{2} + 2k\pi, k\in \mathbb{Z}\}$, for example. $\sin x$ will always reach the value of $1$ in those points.
A: Consider the following theorem: If $\lim_{x\to \infty } f(x)=L$ and $x_n$ is a sequence of numbers such that $\lim _{n\to \infty }x_n =\infty $, then $\lim_{n\to \infty }f(x_n)=L$. If you feel comfortable using this theorem, then you can now prove that $\lim_{x\to \infty }\sin(x)\ne \frac{1}{2}$ by considering a suitable sequence $x_n$ with $\lim_{n\to \infty }x_n=\infty $ and $\sin(x_n)=1$ for all $n$. It then follows that $\lim_{n\to \infty }\sin (x_n)=1\ne \frac{1}{2}$, and thus proves the claim. 
If you haven't seen this theorem before, try proving it. If you want to give a proof that is as close as possible to directly using the definition of limit, then try to figure out why the above works. Choose $\epsilon = \frac{1}{4}$ for instance, and show that for all $M$ there exists $x>M$ with distance between $\sin(x)$ and $\frac{1}{2}$ bigger than $\epsilon $.
