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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.

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1 Answer 1

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The first assertion is indeed true. The second one is false. Take the sine function, for instance.

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  • $\begingroup$ oh thank you, such a simple example, i didn't think of it! $\endgroup$
    – ymir
    Commented Mar 19, 2021 at 17:09
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ Commented Mar 19, 2021 at 18:11
  • $\begingroup$ It's actually possible to have $f(\{M,\infty)\})=\mathbb R$ for every $M>0.$ $\endgroup$
    – zhw.
    Commented Mar 19, 2021 at 20:22
  • $\begingroup$ @JoséCarlosSantos You might want to wait a day or so before saying this. $\endgroup$
    – zhw.
    Commented Mar 19, 2021 at 20:23

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