There exists $\displaystyle\lim_{x\to+\infty}h(\sin x)$ if and only if there is a $c\in\mathbb R$ such that for every $x\in[-1,1]$ we have $h(x)=c$.

Question

Given a function $h\colon\mathbb R\longrightarrow\mathbb R$ show that there is $\displaystyle\lim_{x\to+\infty}h(\sin x)$ if and only if there is a $c\in\mathbb R$ such that for every $x\in[-1,1]$ we have $h(x)=c$.

My Draft

I know that $\lim_{x\to+\infty}h(\sin x)$ exists, so there are three possibilities. $\lim_{x\to+\infty}h(\sin x)=\pm\infty$ or $\lim_{x\to+\infty}h(\sin x)=l, \,l\in\mathbb R$.

Let's deal with this last case. Have $$\lim_{x\to+\infty}h(\sin x)=l, \,l \in\mathbb R \iff \forall \epsilon>0\,\exists\, \delta>0: x>\delta\implies|h(\sin x)-l|<\epsilon.$$

That is, there is a $\delta$ for which if $x>\delta$ the function $h(\sin x)$ is in $]l-\epsilon,l+\epsilon[$.

We want to prove that there is $c$ in $\mathbb R$ such that $\forall x\in[-1,1] \, h(x)=c$.

Suppose $\exists \gamma\in[-1,1]$ such that $h(\gamma)=d\ne c$ and $\forall x\in[-1,1]\backslash\{\gamma \}, \, h(x)=c$.

Like $\gamma\in[-1,1]$, there is $\theta$ such that $\sin\theta=\gamma$. Thus, $h(\gamma)=h(\sin\theta)=d$, for some $\theta$ (this is because the function $\sin$ is "surjective on $[-1,1]$".

Considering, for example, $-\gamma$, as $-\gamma\in[-1,1]$, there is $\alpha$ such that $\sin\alpha=-\gamma$. So $h(-\gamma)=h(\sin\alpha)=c$.

I did this and I don't know how to get out of here or that what I've written so far is $100\%$ correct, even if it's not even relevant to the answer in question. Thank you in advance for the help you have been able to provide me.

Answer 1

You're idea is great, using the surjectivity of $\sin$, but your proof is incomplete. There is also no need to speak about continuity here. Let's use your idea to prove it.

We start by proving that if $\lim_{x\to\infty}h(\sin x)$ exists (and is finite), then $h$ is constant on $[-1,1]$.

Suppose for a contradiction that the limit exists, but $h$ is not constant on $[-1,1]$. Let $\alpha,\beta\in\mathbb{R}$ be two distinct values (i.e. $\alpha\neq\beta$) such that $h(a)=\alpha$ and $h(b)=\beta$ for some $a,b\in[-1,1]$. Now as $\sin$ is surjective, we can find some $\xi,\zeta\in\mathbb{R}$ such that $\sin\xi=a$ and $\sin\zeta=b$. Now consider the sequences $\{x_j\}_{j\in\mathbb{N}}$ and $\{y_j\}_{j\in\mathbb{N}}$ given by

$$x_j=\xi+2\pi j,\quad y_j=\zeta+2\pi j,$$

for which

$$\lim_{j\to\infty}x_j=\infty,\quad \lim_{j\to\infty}y_j=\infty.$$

Notice then than

$$\lim_{j\to\infty}h(\sin x_j)=\lim_{j\to\infty}h(\sin\xi)=\alpha,$$

$$\lim_{j\to\infty}h(\sin y_j)=\lim_{j\to\infty}h(\sin\zeta)=\beta.$$

But since $\alpha\neq\beta$, this contradicts that the limit exists, and the above proposition follows.

Now the other direction of the proof is trivial, as if $h(x)=c$ for all $x\in[-1,1]$, then

$$\lim_{x\to\infty}h(\sin x)=\lim_{x\to\infty}c=c,$$

and so the limit exists.