Let $f\colon (0,1)\to\Bbb R$ be uniformly continous. Prove that $\lim_{n \to\infty}f(x_n)$ exists.

Please verify that my proof is correct.

Problem: Let $$f\colon (0,1)\to\Bbb R$$ be uniformly continous. Let $$(x_n)_{n=1}^\infty$$ be a sequence in (0,1). Suppose that $$\lim_{n \to\infty} x_n=0$$. Prove that $$\lim_{n \to\infty}f(x_n)$$ exists.

My solution: Since $$\lim_{n \to\infty} x_n$$ exists and is equal to $$0$$, this means the sequence $${x_n}$$ is convergent. We know that every convergent sequence of real numbers is cauchy. Therefore $${x_n}$$ is a cauchy sequence. Since $$f$$ is uniformly continous on $$(0,1)$$, for any given $$\epsilon>0$$ , $$\exists\delta>0$$ such that for every pair of points $$x,y\in(0,1)$$ satisfying $$|x-y|<\delta\to|f(x)-f(y)|<\epsilon$$ Since $${x_n}$$ is a cauchy sequence, $$\existsk\in\mathbb{N}$$ such that $$|x_n-x_m|<\delta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall n,m$$
which follows that $$\forall n,m>k$$ $$|f(x_n)-f(x_m)|<\epsilon\:\:\:\:\text{in}\:\:\mathbb{R} \:\:\implies f(x_n) ~~\text{is a Cauchy sequence}$$ Since every Cauchy sequence of $$\mathbb{R}$$ is convergent. Therefore, $$f(x_n)$$ is convergent. Hence $$\lim_{n \to\infty}f(x_n)$$ exists  Does my proof look correct?

• You mean prove $\lim\limits_{n\to \infty}f(x_n)$ exists? Assuming yes, apart from some phrasing nitpicks I have, the proof is right. Commented Apr 26, 2021 at 22:19
• A uniformly continuous function converts a Cauchy sequence into a Cauchy sequence. Commented Apr 26, 2021 at 22:34