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Let $f: [0, \infty) \to \mathbb R$ be a continuous function.

We say that $g: [0, 1] \to [-\infty, \infty]$ is an infinity limit of $f$ if there exists a sequence of positive reals $r_n$ monotonically increasing to $\infty$ such that $f_n: [0, 1] \to [-\infty, \infty]$ defined by $f_n (x) := f(r_n x)$ converges pointwise to $g$, and the convergence is uniform on $[\epsilon, 1]$ for every $\epsilon > 0$.

For example, a function that approaches a limit $L$ at infinity has the constant function $L$ as it’s only infinity limit.

For a continuous function $f$, is the set of infinity limits of $f$ at most countably infinite?

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The answer is no. Indeed, let $$f(x):=\sin\max(0,\ln x)$$ for all real $x>0$, with $f(0):=0$. For any $a\in[0,2\pi)$ and all natural $n$, let $$r_n:=r_{n,a}:=e^{a+2\pi n}.$$ Then $$f(r_n x)=\sin(a+\ln x)$$ for all natural $n$ and all $x\in[1/r_n,1]$. So (as $n\to\infty$), $$f(r_n x)\to\sin(a+\ln x)$$ uniformly in $x\in[\epsilon,1]$ for any real $\epsilon>0$ (actually, uniformly in all $x\in[\epsilon,\infty)$). Also, $f(r_n0)=0\to0$. So, for each $a\in[0,2\pi)$, the function $g_a$ given by $g_a(x):=\sin(a+\ln x)$ for $x\in(0,1]$, with $g_a(0):=0$, is an infinity limit of $f$.

For any distinct $a$ and $b$ in $[0,2\pi)$, we have $g_a\ne g_b$. So, $f$ has unaccountably many infinity limits.


N.B.: You wrote: "For example, a function that approaches a limit $L$ at infinity has the constant function $L$ as it’s only infinity limit." Actually, even if the limit $L:=f(\infty-)$ exists, then in general the infinity limit of $f$ equals $L$ only on $(0,1]$, rather than on $[0,1]$.

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