Prove that $\lim _{x \to \infty} \sin x$ doesn't exist (using delta epsilon)

Question

though there is a question already asked in this site similar to this i want to prove that $\lim _{x \to \infty} \sin x$ doesn't exist using epsilon and delta. I don't know how to do this because sin(${\infty}$) can't be determined. someone please help me in this proof

Answer 1

In addition to the response given by Belgi, here a version with your epsilon but missing a delta ;)

A function $f:\mathbb{R} \to \mathbb{R} $ is said to have a limit $L$ at infinity if for all $\epsilon>0$ there exists a $c\in\mathbb{R}$ so that for all $x>c$ you have $\left| f(x)-L \right|<\epsilon$. Now the negation of this is:

There exists an $\epsilon>0$ so that for all $c\in\mathbb{R}$ there is an $x>c$ with $\left| f(x)-L\right|\geq\epsilon$

Now in your case for any $L\in [0,1)$ (the cases $L=1$ and $L\in [-1,0)$ are similar) you can choose $\epsilon=\frac{1-L}{2}$. Then for any $c\in\mathbb{R}$ set $x=\frac{\pi}{2} + 2\pi \lceil c \rceil$ and get $\left| \sin(x)-L\right|\geq\epsilon$.

Answer 2

Hint :find two subsequences, one that is always zero and the other is always one