Please read the question before you downvote or report as duplicate.
I saw 2 similar instances of this question but I feel that the answers in those questions are incorrect for my question. I don't have enough reputation to comment on those questions so I had to make a new question.
So this is my question:
Let $I$ be an open interval and let $a \in I$. Suppose $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$
Prove or disprove : $f(x) < g(x)$, $\forall x \in I$ $\implies$ $L<M$
All the answers on this site says that the above proposition is false and that $L \le M$.
And they've used construction proofs to show that there $\exists$ $f(x) < g(x)$ such that $L = M$.
But I think the above proposition is true because all the answers on this site ignore the 2 facts, $\forall x \in I$, $f(x) < g(x)$ and $a \in I$.
They've used instances where either $a \notin I$ or instances where $f(x) = g(x)$.
Some of them has used instances where $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \infty$.
But doesn't $\forall x \in I$, $f(x) < g(x)$ imply that this inequality hold for all $x$ values in $I$ and that $f(x)$ and $g(x)$ must be real numbers.(Not $\infty$)
Whether or not $a \in I$ is irrelevant: the value of $L = \lim_{x \to a} f(x)$ does not depend on $f(a)$ and $M = \lim_{x \to a} g(x)$ does not depend on $g(a)$. So, in case $a \in I$, the fact that $f(a) < g(a)$ does tell you nothing at all about about the relation between $L$ and $M$.
For example: take $f(x) = 0$ for $x \neq 0$ and $f(0) = -1$ and take $g(x) = x^2$. Then $f(x) < g(x)$ for all $x$, even for $x = 0$ because I've given $f(0)$ a special value. Still $\lim_{x \to 0} f(x) = 0$ and $\lim_{x \to 0} g(x) = 0$.
If you assume $a \in I$ and assume $f$ and $g$ to be continuous, then of course $L = f(a)$ and $M = g(a)$. The condition $\forall x \in I: f(x) < g(x)$, then gives $f(a) < g(a)$, so $L < M$.