doubt in Proof if f(x) is continuous on [a,b],strictly increasing on (a,b),then f(x) is strictly increasing on [a,b]

Question

This doubt is related to an answer that Paramanand Singh wrote here. The related question was

Prove that if $f(x)$ is continuous on $[a,b]$ ,strictly increasing on $(a,b)$, then $f(x)$ is strictly increasing on $[a,b]$

The related answer was

The idea is very simple. Since $f$ is strictly increasing on $(a, b)$ it follows that if we have $a < x < b$ then we can choose $y, z$ such that $a < y < z < x < b$ and then $f(y) < f(z) < f(x)$. Letting $y \to a^{+}$ we get $f(a) \leq f(z) < f(x)$ and thus $f(a) < f(x)$. Similarly it can be proved that $f(x) < f(b)$. It follows that $f$ is strictly increasing on $[a, b]$.

My doubt: When letting $y \to a^{+}$why do we get $f(a) \leq f(z) < f(x)$
and not $f(a) < f(z) < f(x)$

Answer 1

That's because limits do not preserve strict inequalities like $<$ and $>$. Limits only preserve $\leq$ and $\geq$. So the moment we take limits, strict inequalities turn into non-strict inequalities (or, rather, we weaken our initial hypothesis to use non-strict inequalities, and then we take limits, which preserve such inequalities).

For instance, for any $x\in (0,1)$, we have $0<x^2$. However, taking the limit towards $0$, we get $0\not <\lim_{x\to 0^+} x^2$.