1. ## uniform continuity

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 $\displaystyle [a,b)$.

2. 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:
Use the sequential definition of continuity, i.e. $\displaystyle f$ is continuous if and only if $\displaystyle x_n \rightarrow x$ implies $\displaystyle f(x_n)\rightarrow f(x)$
.

Conversely if the function is uniformly continuous then a straightforward use of the definition shows the limit exists.

Spoiler:
Let $\displaystyle \epsilon>0$ be given and let $\displaystyle x_n$ be any sequence converging to $\displaystyle x$. Then there exists $\displaystyle \delta>0$ such that $\displaystyle |x-y|<\delta$ implies $\displaystyle |f(x)-f(y)|<\epsilon$. Use the Cauchy criterion.