Which of the following are true about sequences?

Question

If $(x_n)$ is a sequence of real numbers such that for every $n$ we have $0<x_n<\frac{1}{n}$ then which of the following is true?

$1.\lim_{n\to\infty}x_n=0$

$2.$If $f$ is continuous function from $\\(0,1)$ then $\\(f(x_n))$ is a cauchy sequence.

$3.$If $g$ is uniformly continuous function on $(0,1)$ then $g(x_n)$ is convergent.

First we can prove by squeeze theorem and for second option we have counter example $\\A=(0,1/n),f(x)=\frac{1}{x},x_n=1-\frac{1}{2n}$

About other options I have no idea.

Answer 1

3. Let $\epsilon > 0$ be arbitrary and let $\delta > 0$ be such that for all $a,b \in (0,1)$ and $|a-b|< \delta$, then $|g(a) - g(b)| < \epsilon$. Since $\{x_n\}_{n=1}^{\infty}$ converges, it is Cauchy, and therefore we can find some $N \in \Bbb{N}$ large enough such that for all $n, m \geq N$, it holds that $|x_n - x_m| < \delta$. Now, for all $n, m \geq N$, $|g(x_n) - g(x_m)| < \epsilon$ since $x_n, x_m \in (0,1)$ with $|x_n - x_m| < \delta$. This shows that $\{g(x_n)\}_{n=1}^{\infty}$ is Cauchy and therefore...