Assume that $f: \mathbb{R} -\{0\}\to \mathbb{R}$ is uniformly continuous. Assume $(x_n)_{n\in\mathbb{N}}\in(\mathbb{R}-\{0\})^\mathbb{N}$ and $(y_n)_{n\in\mathbb{N}}\in(\mathbb{R}-\{0\})^\mathbb{N}$ are both sequences that converge to zero. Show that $(f(x))_{n\in\mathbb{N}}$ and $(f(y))_{n\in\mathbb{N}}$ have the same limit.

To start, I proved that $(f(x))_{n\in\mathbb{N}}$ and $(f(y))_{n\in\mathbb{N}}$ are Cauchy sequences (which converge). But I'm not sure how to proceed. Heuristically, I want to show that the distance between $(f(x))_{n\in\mathbb{N}}$ and $(f(y))_{n\in\mathbb{N}}$ is very small.

My attempt:

Since $f$ is uniformly continuous, then for all $\epsilon>0$ there exists $\delta>0$ such that if $|x-y|<\delta_\epsilon$ then $|f(x)-f(y)|<\epsilon.$

Let's say for all $\eta>0$, if $n>N_1$ then $|x_n|<\eta$ and if $n>N_2$ then $|y_n|<\eta$.

Pick $N=max(N_1,N_2)$ such that when $n>N$ one has $$|x_n-y_n|\leq |x_n|+|y_n|<\eta/2+\eta/2=\eta.$$

But how do I make the leap to concluding something about $(f(x))_{n\in\mathbb{N}}$ and $(f(y))_{n\in\mathbb{N}}$?


Your $\eta$ is arbitrary. Choose it such that $\eta = \eta_\epsilon < \delta_\epsilon$.

Then by the uniform continuity $|f(x_n)-f(y_n)|<\epsilon$, for all $n>N$.


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