Here is the proof of aformentioned theorem.
[MATH] f : Df \rightarrow R, \forall x\in Df x<a, a\in D'f, Df \subset R, f [/MATH] - monotonic function. Then
[MATH]1) \exists \lim_{x \to a }f(x)[/MATH]
2) [MATH]f - increasing \Rightarrow lim_{x \to a}f(x)=sup<f(x)|x \in Df>[/MATH]
3)[MATH]f - decreasing \Rightarrow lim_{x \to a}f(x)=inf<f(x)|x\in Df>[/MATH]
Case when f - increasing, [MATH]supRf \in R, a\in R, x<a[/MATH].
Proof: [MATH]b:=sup<f(x)|x\in Df>[/MATH]
[MATH]b \in R \Rightarrow \forall x \in Df \ \ f(x)\le b[/MATH]
Does [MATH]lim_{x \to a}f(x)=b?[/MATH]
Does [MATH]\forall \epsilon > 0 \ \ \exists \delta >0: \forall x \in Df \ \ x\in (a - \delta; a + \delta) \Rightarrow f(x) \in (b - \epsilon; b+ \epsilon)?[/MATH] Because [MATH]\forall x\in Df \ \ x<a \ \ x\in (a-\delta;a)[/MATH] and because [MATH]b - sup., f(x)\notin (b;b+\epsilon) [/MATH]. Does [MATH]f(x) \in (b-\epsilon;b)?[/MATH]Take a point [MATH]x'[/MATH]. Then if [MATH]\epsilon>0[/MATH] given [MATH]b=sup<f(x)|x\in Df> \ \Rightarrow \exists x'\in Df : f(x')>b-\epsilon [/MATH]Then if [MATH]x'<x \Rightarrow f(x')\le f(x)[/MATH]
[MATH]x' \in (a-\delta;a)\ \ x'>a-\delta [/MATH] then [MATH]\delta=a-x'[/MATH], t.i., [MATH]\delta[/MATH] exists.
I don't understand why [MATH]x'\in (a-\delta;a)?[/MATH] why it is important to note, that if [MATH]x'<x \Rightarrow f(x')\le f(x)[/MATH], how does it help to prove that [MATH]x'\in(a-\delta;a)?[/MATH] Also, the statement that [MATH]x \in (a-\delta;a)[/MATH] should not be proved anyway that such x exists? We can just take it as given?
[MATH] f : Df \rightarrow R, \forall x\in Df x<a, a\in D'f, Df \subset R, f [/MATH] - monotonic function. Then
[MATH]1) \exists \lim_{x \to a }f(x)[/MATH]
2) [MATH]f - increasing \Rightarrow lim_{x \to a}f(x)=sup<f(x)|x \in Df>[/MATH]
3)[MATH]f - decreasing \Rightarrow lim_{x \to a}f(x)=inf<f(x)|x\in Df>[/MATH]
Case when f - increasing, [MATH]supRf \in R, a\in R, x<a[/MATH].
Proof: [MATH]b:=sup<f(x)|x\in Df>[/MATH]
[MATH]b \in R \Rightarrow \forall x \in Df \ \ f(x)\le b[/MATH]
Does [MATH]lim_{x \to a}f(x)=b?[/MATH]
Does [MATH]\forall \epsilon > 0 \ \ \exists \delta >0: \forall x \in Df \ \ x\in (a - \delta; a + \delta) \Rightarrow f(x) \in (b - \epsilon; b+ \epsilon)?[/MATH] Because [MATH]\forall x\in Df \ \ x<a \ \ x\in (a-\delta;a)[/MATH] and because [MATH]b - sup., f(x)\notin (b;b+\epsilon) [/MATH]. Does [MATH]f(x) \in (b-\epsilon;b)?[/MATH]Take a point [MATH]x'[/MATH]. Then if [MATH]\epsilon>0[/MATH] given [MATH]b=sup<f(x)|x\in Df> \ \Rightarrow \exists x'\in Df : f(x')>b-\epsilon [/MATH]Then if [MATH]x'<x \Rightarrow f(x')\le f(x)[/MATH]
[MATH]x' \in (a-\delta;a)\ \ x'>a-\delta [/MATH] then [MATH]\delta=a-x'[/MATH], t.i., [MATH]\delta[/MATH] exists.
I don't understand why [MATH]x'\in (a-\delta;a)?[/MATH] why it is important to note, that if [MATH]x'<x \Rightarrow f(x')\le f(x)[/MATH], how does it help to prove that [MATH]x'\in(a-\delta;a)?[/MATH] Also, the statement that [MATH]x \in (a-\delta;a)[/MATH] should not be proved anyway that such x exists? We can just take it as given?