Let $ f:I \to \mathbb{R} , I=(0,1) $ be uniformly continuous. Then exists $ \lim_{n\to\infty} f(\frac{1}{n}) $

Question

True.

Since $f$ is continuous (because all uniformly continuous function is continuous), we can assume:

$$ f\left(\lim_{n\to\infty} \frac{1}{n}\right) $$

Since $ \lim_{n\to\infty} \frac{1}{n} $ is bounded in $ (0,1] $ and $ I \subset (0,1] $, we have by hypothesis $f$ uniformly continuous then

$$ \lim_{n\to\infty} f\left(\frac{1}{n}\right) \text{, exists.} $$

Answer 1

Okay, as David says: To prove your claim it is sufficient to prove that the sequence $(f(1/n))_{n=1}^\infty$ is Cauchy. Completeness of $\mathbb R$ will then give you the desired convergence. To see that the sequence is Cauchy let $\epsilon > 0$ and note that, since $f$ is uniformly continuous, there exists a $\delta > 0$ such that $\lvert f(x)-f(y) \rvert < \epsilon$, whenever $\lvert x-y \rvert < \delta$. For $m,n$ big enough (such that $\lvert 1/n - 1/m \rvert = \lvert (m-n)/(mn) \rvert < \delta$), it follows that $$ \lvert f(1/n) - f(1/m) \rvert < \epsilon $$ for large enough $m,n$.

Answer 2

Hint: You are trying to prove that $f$ can be extended to $[0,1)$ by putting $f(0)=\lim f(\frac{1}{n})$.

By uniform continuity of $f$ you can prove that limit exists (how?), and therefore $f$ can be extended in a coherent way (a way that makes that $f$ continuous in $[0,1)$ )