I would like to find a "counterexample" for the following false statement.
Let $f$ be continuous on $[a, \infty)$ where $a \in \mathbb{R}$ and the codomain is also $\mathbb{R}$. If $f$ is uniformly continuous and bounded on $[a, \infty)$, then $\lim _{x\to \infty}$ $f(x)$ exists and is finite.
I initially thought of $f(x)=x$ where the domain is $[a, \infty)$ and $a \in \mathbb{R}$. Although $f$ is uniformly continuous on $[a, \infty)$, $\lim _{x\to \infty}$ $f(x)$ is not finite. However, it is not bounded on $[a, \infty)$, which is another condition of this problem. So, I thought I should find the right counterexample. I would very much appreciate your help.
Use $f(x) = \sin{x}$.
It is bounded, but $\lim_{x \rightarrow \infty} f(x)$ does not exist.
EDIT
However you can look at the $\limsup$ (if you know this concept). It is defined as follows $$ \limsup_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \sup_{y \geq x} f(y). $$ Since $ \sup_{y \geq x} f(y)$ decreases with increasing $x$ and $f$ is bounded, this function will have a limit at infinity, this is the $\limsup$ of $f$.
You can do the same if you define $$ \liminf_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \inf_{y \geq x} f(y). $$