Let $\displaystyle f$ be continuous on $\displaystyle [a,b)$. Prove that$\displaystyle lim_{x \rightarrow b^-}f(x)$ exists $\displaystyle iff$ $\displaystyle f$ is uniformly continuous on $\displaystyle [a,b)$.
If we know $\displaystyle \displaystyle\lim_{x\rightarrow b-}f(x)$ exists and we want to show f is actually uniformly continuous, it helps to invoke a nice theorem.
Theorem: If $\displaystyle f$ is a continuous real function on a closed and bounded set of real numbers, then $\displaystyle f$ is in fact uniformly continuous.
Then if we can define a value of $\displaystyle f$ at $\displaystyle b$ which maintains continuity, we've got it!
.Spoiler:
Conversely if the function is uniformly continuous then a straightforward use of the definition shows the limit exists.
Spoiler: