I just used the Heine's definition.
Let $\alpha,\delta \in \mathbb{R}$ such that $\sin(\alpha)=a$ and $\sin(\delta)=b$. Let $(u_{n})=\alpha+2\pi n$ and $(v_{n})=\delta+2\pi n$ and $f(x)=\sin(x)$.
So one have,
$$\lim\limits_{n\rightarrow \infty} u_{n}=+\infty$$
$$\lim\limits_{n\rightarrow \infty} v_{n}=+\infty$$
But,
$\lim\limits_{n\rightarrow \infty} f(u_{n})=a$
$\lim\limits_{n\rightarrow \infty} f(v_{n})=b$
Because these last limits aren't equal, the sine function don't have limit to infinity. Is this proof correct? Thanks.
Best Answer
Another proof would be to note that $\lim_{x \to \infty} \left( \sin(\frac{\pi}{2}+2\lfloor x \rfloor \pi) - \sin ( 2\lfloor x \rfloor \pi) \right) = 1$, hence $\sin$ has no limit as $x \to \infty$. (If $\lim_{x \to \infty} f(x) = L$, then $\lim_{x \to \infty} \left( f(x+y)-f(x) \right) = 0$ for all $y$).