Is this statement about limits true? Edit: $f$ is differentiable on $ℝ$
If we know that:
$∀a,b,a<b⇒f(a)<f(b)$
Can we say that
$\lim_{x \to a}f(x)<f(b)$
If so, how would you prove it?
 A: Since $f$ is differentiable, then it is continuous. Then we have, for $a<b$:
$$\lim_{x\to a}f(x)=f(a)<f(b)$$
Is that what you need?
A: If $f$ is differentiable on $\mathbb{R}$ then $f$ is continuous on $\mathbb{R}$. So,
$$
\lim_{x\rightarrow a}f(x)=f(a)<f(b).
$$
A: Ok here is a slight generalization because the problem is trivial if $f$ is continuous.

Theorem: Let $f:\mathbb{R} \to\mathbb {R} $ be such that for all $a, b\in \mathbb {R} $ with $a<b$ we have $f(a) <f(b) $. Such a function is said to be strictly increasing. Then for any $a,b\in\mathbb {R} $ with $a<b$ we have $$\lim_{x\to a^{-}} f(x) \leq f(a) \leq \lim_{x\to a^{+}} f(x) <f(b) $$

The above result does not require any conditions like continuity or differentiability. Just the increasing nature of function is required. If the function is strictly decreasing then the inequalities in the conclusion get reversed. Also if we remove the word "strictly" from the hypotheses then the last inequality related with $f(b) $ gets replaced by the weaker version.
To prove the theorem note that if $x<a$ then $f(x) <f(a) $ and thus as $x$ increases and remains less than $a$, $f(x) $ also increases and remains less than $f(a)$. Thus the set $$A=\{f(x) \mid x<a\} $$ is non-empty and bounded above by $f(a) $. Therefore $s=\sup A$ exists and since $f(a) $ is an upper bound for $A$ we have $s\leq f(a) $. Let $\epsilon>0$ then by definition of supremum we have some $x'<a$ and $f(x')>s-\epsilon$. Let $\delta =a-x'>0$ and then for all $x$ with $x'=a-\delta <x<a$ we have $$s-\epsilon <f(x') <f(x) \leq s<s+\epsilon$$ Then by definition of limit $\lim_{x\to a^{-}} f(x) =s\leq f(a) $. And by similar argument $\lim_{x\to a^{+}} f(x) $ exists and $f(a) $ does not exceed it.
Next note that since $a<b$ we can find a $c$ such that $a<c<b$. And if $a<x<c$ then $f(a) <f(x) <f(c) $ so that by taking limits as $x\to a^{+} $ we have $\lim_{x\to a^{+}} f(x) \leq f(c) <f(b) $ and we are done. 
