Let $g : (a, b) → R$ be uniformly continuous on $(a, b)$. Let $\{x_n\}_{n=1}^\infty$ be a sequence in $(a, b)$ converging to $a$. Prove that $\{g(x_n)\}_{n=1}^\infty$ converges.
The general idea here is to use the uniform continuity of g as well as the fact that since the $\{x_n\}$ converges it is Cauchy and to prove that $\{g(x_n)\}_{n=1}^\infty$ is Cauchy.
From uniform continuity we have $\forall\epsilon > 0, \exists\delta > 0$ such that $|x - y| < \delta$ with $x,y \in (a,b)$ then $|g(x) - g(y)| < \epsilon$.
And since $\{x_n\}_{n=1}^\infty$ is Cauchy we have $\forall\epsilon > 0, \exists N \in J$ such that $\forall n,m \ge N$ then $|x_m - x_n| < \epsilon$.
However, I'm not sure how to combine these two given properties in order to get the Cauchy property for $\{g(x_n)\}_{n=1}^\infty$.
Thanks for the help!
Let $\varepsilon > 0$. Because $g$ is uniformly continuous there exists $\delta > 0$ s.t. $$ |x-y| < \delta \implies |g(x)-g(y)| < \varepsilon, \; x,y, \in (a,b) $$ Because $(x_n)_{n=1}^\infty$ is Cauchy sequence, there exists $n_0 \in \Bbb{N}$ s.t. $|x_n - x_k| < \delta$ when $n \geq n_0$ and $k \geq n_0$. So, if $n, k \geq n_0$ holds, we have $$ |g(x_n) - g(x_k)| < \varepsilon $$ so $(g(x_n))_{n=1}^\infty$ is a Cauchy sequence. Because $\Bbb{R}$ is complete, $(g(x_n))_{n=1}^\infty$ converges.