Check my proof that a monotone function has one-sided limits at any point

Question

Here $f:\mathbb{R} \rightarrow \mathbb{R} $.
We consider the case when $f$ is an increasing function.
Let $A=\{f(x) | x<x_0\}$. $A$ is bounded from above by $f(x_0)$,so there exists $\alpha \in \mathbb{R} $ so that $\sup A= \alpha$.
Let $\epsilon >0$. Since $\alpha - \epsilon < \alpha => \exists x_1 \in \mathbb{R}, x_1<x_0$ so that $f(x_1)>\alpha - \epsilon$.
$=>\forall x\in (x_1, x_0), \alpha - \epsilon < f(x_1) \le f(x) <\alpha + \epsilon$
Let $\delta_1=x_0 - x_1$.Then $\forall x\in \mathbb{R}, |x-x_0|<\delta_1,|f(x)-\alpha|<\epsilon$,so $\lim_{x\to x_0^{-}} f(x) =\alpha$.
Let $B=\{f(x) | x>x_0\}$. $B$ is bounded from below by $f(x_0)$,so there exists $\beta \in \mathbb{R} $ so that $\inf B= \beta$.
Let $\epsilon_1 >0$. Since $\beta + \epsilon_1 > \beta => \exists x_2 \in \mathbb{R}, x_2>x_0$ so that $f(x_2)<\beta + \epsilon_1$.
$=>\forall x\in (x_0, x_2), \beta + \epsilon_1 > f(x_2) \ge f(x) >\beta - \epsilon_1$
Let $\delta_2=x_2 - x_0$.Then $\forall x\in \mathbb{R}, |x-x_0|<\delta_2,|f(x)-\beta|<\epsilon_1$,so $\lim_{x\to x_0^{+}} f(x) =\beta$.

Answer 1

It is really good, there is just one small mistake. When you defined $\delta_1$ you wrote the required inequality $|f(x)-\alpha|<\epsilon$ for all $x\in\mathbb{R}$ such that $|x-x_0|<\delta_1$. But the inequality might not hold if $x>x_0$. This is a one sided limit so you needed to write for all $x\in\mathbb{R}$ such that $x_0-\delta_1<x<x_0$. Same thing when you defined $\delta_2$.