I try to prove a simple statement:
Let $a,b\in R$, $a<b$, and $F:(a,b)\rightarrow R$. Given $F(x)<A$ for all $x\in (a,b)$, we have $lim_{x\rightarrow c} F(x)\leq A$ for $c=a,b$ (assuming limits exist at endpoints).
Here is my proof:
Let $c=a$. Suppose $lim_{x\rightarrow a} F(x)> A$. Choose $\epsilon= lim_{x\rightarrow a} F(x)-A$. Then, there exists $\delta>0$ such that for all $x$ in the punctured neighborhood $N_{\delta}^* (a)\cap (a,b)$, we have $F(x)>A$, contradicting the fact that $F(x)<A$ for all $x\in (a,b)$.
Could someone verify my proof? I am not confident, because it seems fairly short. Also, is there any way to emphasize the equality can be true in limits? Thanks in advance!
You start with a function defined on $(a,b)$. If $c \in [a,b]$ and $\lim_{x\rightarrow c} F(x)$ exists, then your argument is correct without any other assumption on $F$.
So let us see what we can say about the existence of the limit.
(1) If $c \in (a,b)$ and $F$ is continuous in $c$, then the limit exists. In this case $\lim_{x\rightarrow c} F(x) = F(c) < A$.
(2) The limit may also exist for functions which are not continuous in $c$. Take for example $F : (0,2) \to \mathbb{R}$, $F(x) = x$ for $x < 1$, $F(1) = 0$ and $F(x) = 2 - x$ for $x > 1$. Then $F(x) < 1$ for all $x$ and $\lim_{x\rightarrow 1} F(x) = 1$.
(3) For $c = a, b$ the existence of the limit does not follow from continuity or differentiability properties of $F$. Consider for example $F : (0,1) \to \mathbb{R}, F(x) = \sin \frac{1}{x}$. Then $\lim_{x\rightarrow 1} F(x)$ exists (it is $= \sin 1$), but $\lim_{x\rightarrow 0} F(x)$ does not exist.