Limit of $f(\sin x)$ doesn't exist I was given the following question:
Let $f : [-1,1] \to \mathbb R$ be a function. Suppose that $f$ is not a constant function. Define the function $g : \mathbb R \to \mathbb R$ by
$$g(x) = f(\sin x).$$
Prove that the limit $\lim \limits_{x \to \infty}g(x)$ doesn't exist.
Now, I do know that I need to start by assuming towards contradiction that the limit does exist. And I do understand that $\lim \limits_{x \to \infty}\sin x$ doesn't exist, but I can't find a way to contradict this, by choosing the right $\epsilon$ and $x$, since I barely know anything about $f$.
 A: Let $a,b, f(a)$ is distinct of $f(b)$, $x=arcsin(a), y=arcsin(b)$
$x_n=arcsin(a)+2n\pi, y_n=y+2n\pi$, $g(x_n)=f(a), g(y_n)=f(b)$,
$lim_{n\rightarrow +\infty}x_n=lim_{n\rightarrow +\infty}y_n=+\infty$ implies that $lim_{z\rightarrow +\infty}g(z)$ does not exist.
A: Suppose that $\lim_{x\to\infty}f\bigl(\sin(x)\bigr)=l$, for some $l\in\mathbb R$, nd that $f$ is not constant. Now, let $a,b\in\mathbb R$ be two distinct real numbers from the range of $f$. Then $a=f(\alpha)$ and  $b=f(\beta)$, for some $\alpha,\beta\in[-1,1]$ and, since $f(\alpha)\neq f(\beta)$, $\alpha\neq\beta$. But then, since we have $\sin(x)=\alpha$ for arbitrarily large values of $x$ and we have $\sin(x)=\beta$ for arbitrarily large values of $x$ too, if you take $\varepsilon$ small enough so that $(l-\varepsilon,l+\varepsilon)$ doesn't contain both numbers $a$ and $b$, it is false that there is a $M>0$ such that$$x>M\implies f(x)\in(l-\varepsilon,l+\varepsilon).$$But this is impossible, since we are assuming that $\lim_{x\to\infty}f\bigl(\sin(x)\bigr)=l$.
A: Let $a,b\in[-1,1]$ be given such that $f(a)\neq f(b)$. Now, define that sequence
$$x_n=\arcsin(a)+2\pi n$$
$$y_n=\arcsin(b)+2\pi n$$
Clearly both these go to infinity. Then
$$f(\sin(x_n))=f(\sin(\arcsin(a)+2\pi n))=f(\sin(\arcsin(a)))=f(a)$$
$$f(\sin(y_n))=f(\sin(\arcsin(b)+2\pi n))=f(\sin(\arcsin(b)))=f(b)$$
If we assume that $\lim_{x\to\infty}g(x)$ exists and is $L$, then
$$\lim_{n\to\infty}x_n=\infty\Rightarrow \lim_{n\to\infty}g(x_n)=L$$
However
$$f(a)=\lim_{n\to\infty}g(x_n)=\lim_{n\to\infty}g(y_n)=f(b)$$
which is a contradiction. We conclude $\lim_{x\to\infty}g(x)$ does not exist.
