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 $(f(y))_{n\in\mathbb{N}}$?