If $f(x)>g(x)$ for some interval then is it right to say $\lim(f(x))\geq\lim(g(x))$ at the extreme point of that interval? If $f(x) >g(x)$ for $x\in(2,5)$ and $f(x),g(x)$ are defined for all real values of $x$ and $\text{lim}(f(x))$ at $2$ and $\text{lim}(g(x))$ at $2$ exist where $2$ is the extreme point of the interval then is it correct to say $\text{lim}(f(x)) \geq \text{lim}(g(x))$ at $2$?
 A: This is true since you've given that the limits are defined at $x=2$.
This is a limit inequality theorem. Please see the proof here (theorem 2.23):
https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch2.pdf
$g(x) < f(x)$ for $x \in (2,5)$
So we can write
$g(x) \le g(x)$ for $x \in (2,5)$
We can directly apply the the limit inequality theorem theorem.
$\lim_{x \to 2} g(x) \le \lim_{x \to 2} f(x)$
Note that the theorem applies to 'accumulation points' which mean 'limit points' and that includes 2 and 5.
For the infinite case, it is convenient to look at limits of sequences. This theorem here gives a proof that the inequalities apply to the limit at infinity:
https://proofwiki.org/wiki/Inequality_Rule_for_Real_Sequences
EDIT: Note that I've edited since we're not actually using the squeeze theorem here as we're not saying the limits are equal.
A: Since we assume that we can continuously expand $f$ and $g$ to $[a,b]$ (in general, there is at most one continuous extension of a continuous function defined on an open interval, see my proof* below), the question is:
If $f,g\colon[a,b]\to\mathbb{R}$ are continuous and $g(x)<f(x)$ for all $x\in(a,b)$, can we say that $g(x)\leq f(x)$ for all $x\in[a,b]$?
If this was not the case, $f-g$ would not be continuous. But since $f$ and $g$ are continuous functions, $f-g$ is continuous. So the answer is yes.
*Proof:
Let $f,g\colon[a,b]\to\mathbb{R}$ be continuous and $f(x)=g(x)$ for all $x\in(a,b)$. $X:=(f-g)^{-1}(\mathbb{R}\setminus\{0\})$ is open, because $f-g$ is continuous. As $X$ is a subset of $\{a,b\}$, which has the (Lebesgue) measure zero, the measure of $X$ is also zero. Since the empty set is the only open set with measure zero, $X$ has to be the empty set.
