Proof Using Formal Definition of Limit Prove that the limit of sin(x) as it approaches infinity does not exist using the formal definition of a limit.
 A: Let $\epsilon=1$ (or, if you wish, something smaller, like $1/4$). If there was a number $a$ such that $\lim_{x\to\infty} \sin x=a$, then there would be a $B$ such that  $|\sin x-a|\lt \epsilon$ for all $x\gt B$. 
But there are arbitrarily large $s$ such that $\sin s=1$, and also arbitrarily large $t$ such that $\sin t=-1$. Show that $a$ cannot be simultaneously at distance $\lt 1$ from $1$ and from $-1$. Formally, this part of the argument uses the Triangle Inequality.
A: Suppose it does exist and seek a contradiction. It shouldn't be too difficult to find something inconsistent with your proposed limit and the oscillatory nature of sine.
A: Suppose that you use the limit of sequence to define the limit of function.
To prove that $\displaystyle \lim_{x\rightarrow\infty}\sin(x)$ does not exist we choose two different real sequences $\{u_n\}$ and $\{v_n\}$ such that
$$
\lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}v_n=\infty,
$$
and
$$
\lim_{n\rightarrow\infty}\sin(u_n)\ne\lim_{n\rightarrow\infty}\sin(v_n).
$$
For example we choose
$$
u_n=\frac{\pi}{2}+n2\pi, \quad v_n=\pi+n2\pi \quad (n\in \mathbb{N}).
$$
Then
$$
\lim_{n\rightarrow\infty}u_n=\lim_{n\rightarrow\infty}v_n=\infty,
$$
and
$$
\lim_{n\rightarrow\infty}\sin(u_n)=1\ne0=\lim_{n\rightarrow\infty}\sin(v_n).
$$
