composition of homemomerphism and continuous function and limit at $\infty$ Let $h\colon \Bbb R\to \Bbb R$ be a continuous function and $\lim_{x\rightarrow\infty}h(x)=0$  and $\lim_{x\rightarrow -\infty}h(x)=0.$ Let $f\colon (a,b)\to\Bbb R$ be a homeomoerphism. I do not see why $$\lim_{x\rightarrow a^{+}}(h\circ f)(x)=0 \ \  \text{and} \ \  \lim_{x\rightarrow b^{-}}(h\circ f)(x)=0.$$   Does that correct? What does this tell me in terms of cluster point of $h\circ g$, that is, it says $0$ is cluster point of $h\circ f$ at $a$ and $b$ ? Is that right? Any help will be appreciated greatly.
 A: Let $f:(a,b)\to\mathbb{R}$ be a homeomorphism.
Claim: We claim that $\lim\limits_{x\to b^-}f(x)=\infty$ or $\lim\limits_{x\to b^-}f(x)=-\infty$.
Proof: Suppose $\lim\limits_{x\to b^-}f(x)\neq\infty$ and $\lim\limits_{x\to b^-}f(x)\neq-\infty$.
Then, there exists an $M>0$ such that for all $\delta>0$ there exists an $x\in (a,b)$ with $0<b-x<\delta$ but $f(x)\leq M$ (this is the direct negation of $\lim\limits_{x\to b^-}f(x)=\infty$).
Similarly, there exists an $m>0$ such that for all $\delta>0$ there exists an $x\in(a,b)$ with $0<b-x<\delta$ but $-m\leq f(x)$.
Since $f$ is continuous and injective on $(a,b)$ into $\mathbb{R}$, then $f$ is either strictly increasing or strictly decreasing. Suppose first that it is strictly increasing.
Suppose, for contradiction, there exists a $y\in(a,b)$ with $f(y)>M$. Then $b-y>0$ so there exists an $x\in(a,b)$ with $b-x<b-y$ and $f(x)\leq M<f(y)$ but $y<x$, contradicting the fact that $f$ is strictly increasing.
This means we must have $f(x)\leq M$ for all $x\in(a,b)$. But then $M$ is an upper bound for the image of $f$, which is equal to $\mathbb{R}$, and is thus a contradiction, since $\mathbb{R}$ is unbounded. The case when $f$ is strictly decreasing is similar. This completes the proof of the claim.
Since $h:\mathbb{R}\to\mathbb{R}$ is a function with $\lim\limits_{x\to\pm\infty}h(x)=0$ and $\lim\limits_{x\to b^-}f(x)=\pm\infty$, then we must have that $$\lim\limits_{x\to b^-}(h\circ f)(x)=0\tag{$\star$}$$
If anything is unclear, I would be happy to elaborate.
Edit: We want to show that $0$ is a cluster point of $h\circ f$.
Let $U$ be an open neighborhood of $0$ and let $r>0$. Since $U$ is an open neighborhood of $0$, there exists an $\varepsilon>0$ such that $(-\varepsilon,\varepsilon)\subseteq U$. Set $B=(b-r,b+r)$. We want to show that $(h\circ f)^{-1}(U)\cap B\neq\emptyset$. By $(\star)$, there exists a $\delta>0$ such that if $0<b-x<\delta$, then $|(h\circ f)(x)|<\varepsilon$. We can modify $\delta$ to be arbitrarily small if needed so that $b-r<b-\delta<x<b$. Then, for $x\in(a,b)$ with $0<b-x<\delta$, we have $|(h\circ f)(x)|<\varepsilon$, which implies $x\in (h\circ f)^{-1}(U)$, and we also have $x\in(b-r,b+r)$.
