# $(f(x))_{n\in\mathbb{N}}$ and $(f(y))_{n\in\mathbb{N}}$ have the same limit.

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 $$n>N$$.