Is it true that, if $f$ is uniformly continuous in $(a,b),$ then the limits $\lim_{x\to a^+} f(x)$ and $\lim_{x\to b^-} f(x)$ exist? Let $f$ be a continuous function in $(a,b)$.

I know that:
If the limits $\lim_{x\to a^+} f(x)$ and $\lim_{x\to b^-} f(x)$ exist, then $f$ is uniformly continuous in $(a,b)$.

Is the inverse also true?
 A: YES.
If we define
$$
g(x)=\left\{
\begin{array}{ccl}
\lim_{x\to a^+}f(x) & \text{if} & x=a, \\
f(x) & \text{if} & x\in (a,b), \\
\lim_{x\to b^-}f(x) & \text{if} & x=b.
\end{array}
\right.
$$
then $g$ is continuous in $[a,b]$, and hence uniformly continuous.
Consequently, so is
$$
f=g|_{(a,b)}
$$
INVERSE. If $f: (a,b)\to\mathbb R$ is uniformly continuous, then the limits
$$
\lim_{x\to a^+}f(x)\quad\text{and}\quad
\lim_{x\to b^-}f(x)
$$
exist.
Proof. We need the following:
Lemma. If $f$ uniformly continuous and $\{a_n\}$ is a Cauchy sequence, then so is $\{f(a_n)\}$.
Assuming the above, let $x_n\to a$, then $\{f(x_n)\}$ is Cauchy and hence convergent. Let $\lim_{n\to\infty}f(x_n)=\ell$. We shall show that $\lim_{x\to a^+}f(x)=\ell$. If not, then there would be an $\varepsilon>0$ and a sequence $\{y_n\}\subset (a,b)$, such that $y_n\to a$ and $|f(y_n)-\ell|\ge \varepsilon$. Let $\{z_n\}$ be the sequence
$$
z_1=x_1,\, z_2=y_1,\, z_3=x_2,\,z_4=y_2,\,z_5=x_3,\,z_6=y_3,\,
\text{etc},
$$
then $z_n\to a$, hence $\{f(z_n)\}$ is a Cauchy sequence. Since $f(z_{2n-1})=f(x_n)\to\ell$, then $f(z_n)\to\ell$,
and thus $f(z_{2n})=f(y_n)\to\ell$, contradiction.
Note. More generally, if $A\subset\mathbb R$ and $f: A\to\mathbb R$ is uniformly continuous on $A$, then the limit $\lim_{x\to a}f(x)$ exists for every $a\in \overline{A}$, and $f$ extends continuously on $\overline{A}$.
