# Uniform continuity on $(0, \infty)$ and limits of the function at $\lim$ at $0$ and $\infty$

Are the following statements True or False?

1. If $$f : (0, \infty) \rightarrow \mathbb{R}$$ is uniformly continuous then $$\lim_{x \rightarrow 0} f(x)$$ exists.

2. If $$f : (0, \infty) \rightarrow \mathbb{R}$$ is uniformly continuous then $$\lim_{x \rightarrow \infty} f(x)$$ exists.

For part 1, this is what I tried : consider $$\{x_n\}$$, $$\{y_n\}$$ be two Cauchy sequences converging to 0. Then by uniform continuity $$\{f(x_n)\}$$ and $$\{f(y_n)\}$$ are Cauchy and convergent to the same point $$l = \lim_{n \rightarrow \infty} f(x_n)$$. To show $$l = f(\lim_{n \rightarrow \infty} x_n) = f(0)$$, which is satisfied as f is continuous on $$(0, \infty)$$ hence the limits can be exchanged. Hence $$\lim_{x \rightarrow 0} f(x)$$ exists.

For the second part, by uniform continuity on $$[b, \infty)$$, $$b$$ sufficiently large, given $$\epsilon > 0$$, there exists $$\delta > 0$$ such that whenever $$|x-y| < \delta$$, $$|f(x)- f(y)| < \epsilon$$. Consider a sequence $$x_n \rightarrow \infty$$. I am stuck here.

Could you please check my ideas and help me finish the problems? Thank you.

• It's actually possible to have $f(\{M,\infty)\})=\mathbb R$ for every $M>0.$