uniform continuity on $(a, b]$ implies limit at $a^+$ exists and finite

Question

Let a uniformly continuous function $f$ on $(a, b]$. Prove that $\lim_{x\rightarrow a^+} f(x)$ exists and finite.

What I did so far:
from the definition of uniform continuity:

$$\forall\varepsilon >0.\exists\delta>0.\forall x,y\in(a,b]:\left| x-y \right|<\delta \Rightarrow \left| f(x)-f(y) \right| < \varepsilon$$

In particular, the statement is true for the sequence $\varepsilon_n = \frac{1}{n}$ and the interval $(a,a+\delta)$

$$\forall\varepsilon_n.\exists\delta>0.\forall x,y\in (a,a+\delta).\left| f(x) - f(y) \right| < \frac{1}{n}$$

I'm kinda stuck at this point, though I think I'm on the right path.
How to proceed?

Thanks.

Answer 1

You can apply item 2 of this post. It is easy to check that uniform continuity implies the validity of the condition stated at item 2.

Answer 2

Here's a way without Cauchy sequences.

We prove first that $f$ is bounded on $(a,b]$. By uniform continuity, there is some $\eta \in (0,b-a)$ such that $|x-y|<\eta \implies |f(x)-f(y)|<1$.
Since $f$ is continuous on the compact interval $[a+\eta,b]$, it is bounded by some $M$ on this interval. If $x$ is in $(a,a+\eta)$, then $|f(x)|\leq |f(x)-f(a+\eta)|+|f(a+\eta)|\leq 1+M$. Hence $f$ is bounded by $1+M$ on $(a,b]$.

Next, the sequence $(f(a+1/n))_{n\geq 1}$ is bounded, thus by Bolzano–Weierstrass it has a subsequence $(f(a+1/n_k))_{k\geq 1}$ that converges to some $L$. It remains to prove that $\lim_{a^+} f = L$.
Given $\epsilon>0$ there is some $\eta \in (0,b-a)$ such that $|x-y|<\eta \implies |f(x)-f(y)|<\epsilon/2$. We can also find some $N$ such that $\frac 1N < \eta$ and $|f(a+\frac 1N)-L|<\epsilon/2$. Finally, if $x\in (a,a+\eta)$, $$|f(x)-L|\leq |f(x)-f(a+\frac 1N)|+|f(a+\frac 1N)-L|<\epsilon.$$

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