# Infinity limits of functions

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?

## 1 Answer

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 $$[0,1]$$.