$f:Df \rightarrow R, \forall x\in Df x<a, a\in D'f, Df \subset R, f$ - monotonic. Then
$1) \exists \lim_{x \to a }f(x)$
2) $f - increasing \Rightarrow lim_{x \to a}f(x)=sup<f(x)|x \in Df>$
3)$f - decreasing \Rightarrow lim_{x \to a}f(x)=inf<f(x)|x\in Df>$
Case when f - increasing, $supRf \in R, a\in R, x<a$.
Proof: $b:=sup<f(x)|x\in Df>$
$b \in R \Rightarrow \forall x \in Df \ \ f(x)\le b$
Does $lim_{x \to a}f(x)=b?$
Does $\forall \epsilon > 0 \ \ \exists \delta >0: \forall x \in Df x\in (a - \delta; a + \delta) \Rightarrow f(x) \in (b - \epsilon; b+ \epsilon)?$ Because $\forall x\in Df x<a \ \ x\in (a-\delta;a)$ and because $b - sup., f(x)\notin (b;b+\epsilon) $. Does $f(x) \in (b-\epsilon;b)?$ Take a point $x'$. Then if $\epsilon>0$ given $b=sup<f(x)|x\in Df> \ \Rightarrow \exists x'\in Df : f(x')>b-\epsilon $ Then if $x'<x \Rightarrow f(x')\le f(x)$
$x' \in (a-\delta;a)\ \ x'>a-\delta $ then $\delta=a-x'$, t.i., $\delta$ exists.
I don't understand why $x'\in (a-\delta;a)?$ why it is important to note, that if $x'<x \Rightarrow f(x')\le f(x)$, how does it help to prove that $x'\in(a-\delta;a)?$