Limit approaching infinity of sine function

If $\sin x$ had a limit $L$ for $x\to\infty$, then for every sequence $(x_n)$ such that $x_n\to\infty$ we would have $$\lim\limits_{n\to\infty} \sin x_n=L.$$ In particular, this limit would exist and would have the same value for every choice of such sequence $(x_n)$. (See e.g. here http://en.wikipedia.org/wiki/Limit_of_a_function#Sequential_limits ; but this theorem was probably mentioned in your lecture/textbook.)

If you choose $x_n= 2n\pi$, then this limit is equal $0$.

If you choose $x_n=\frac\pi2+2n\pi$, then this limit is equal to $1$.

The set of limit points of $\sin x$ as $x \to \infty$ is $[-1,1]$. In particular, it is not a single point, and thus the limit doesn't exist.