One sided limit of an increasing function defined on an open interval

Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$? If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.

Both statements are true.

There are two cases to consider. Let $S=\{f(x):x\in(a,b)\}$. Suppose first that $S$ is bounded below, i.e., that there is some $y\in\Bbb R$ such that $y<f(x)$ for all $x\in(a,b)$. Then by the completeness of $\Bbb R$ the set $S$ has a greatest lower bound (or infimum) $u$. I claim that $\lim\limits_{x\to a^+}f(x)=u$.

To prove this, let $\epsilon>0$. Then $u+\epsilon>u$, so $u+\epsilon$ is not a lower bound of $S$, and there is therefore some $a+\delta\in(a,b)$ such that $f(a+\delta)<u+\epsilon$. But $f$ is strictly increasing, so $$u\le f(x)<f(a+\delta)<u+\epsilon$$ for every $x\in(a,a+\delta)$, and therefore $\lim\limits_{x\to a^+}f(x)=u$.

The second is an immediate consequence of the argument just given: $u\le f(x)$ for every $x\in(a,b)$.

Now suppose that $S$ is not bounded below. Then for every $u\in\Bbb R$ there is an $x_u\in(a,b)$ such that $f(x_u)<u$. Since $f$ is strictly increasing, we have $f(x)<f(x_u)<u$ for every $x\in(a,x_u)$, and it now follows easily that $\lim\limits_{x\to a^+}f(x)=-\infty$.

As for the second statement, I think there is a more precise result. Below is my trial, please check if it is correct.

Lemma Let $f\colon D\subset\mathbb{R}\to\mathbb{R}.$ Suppose that $a, b\in D$ with $a<b, $ and the point $a$ is a right-sided limit point of $D.$ Let $f$ is strictly increasing on $(a,b)\cap D.$ If $f$ has a right limit $f(a+0)$ at $a,$ then \begin{gather*} f(x)>f(a+0),\qquad\forall x\in (a,b)\cap D. \end{gather*}

Proof: Let $A=f(a+0):=\lim\limits_{x\to a+0}f(x).$
If $A=-\infty,$ then there is nothing to show. So we assume that $A\in\mathbb{R}.$

We prove the statement by contradiction. If otherwise there exists $x_1\in (a,b)\cap D$ such that $f(x_1)<A,$ then, for $\epsilon_1=\frac{A-f(x_1)}{2},$ there exists $\delta_1>0$ with $\delta<x_1-a,$ such that for all $x\in (a,b)\cap D,$ if $0<a-x<\delta_1,$ then \begin{gather*} f(x)>A-\epsilon_1=A-\frac{A-f(x_1)}{2}=\frac{A+f(x_1)}{2}>f(x_1). \end{gather*} Pick $y_1\in (a,a+\delta_1),$ then $a<y_1<a+\delta_1<a+x_1-a=x_1,$ and $f(y_1)>f(x_1).$ But this contradicts that $f$ is strictly increasing on $(a,b)\cap D.$ If otherwise there also exists $x_2\in (a,b)\cap D$ such that $f(x_2)=A,$ then, pick arbitrarily $x_3\in (a,x_2),$ then, by strict increasingness of $f$ on $(a,b)\cap D,$ we have $f(x_3)<f(x_2)=A.$ By what we have proved, a contradiction also occurs. Therefore we have proved that $f(x)>f(a+0),$ for all $x\in (a,b)\cap D.$