Let $f:[a,b) \to \mathbb{R}$ be a strictly monotone increasing continuous function on a half closed interval $[a,b)$, and let $d$ be a real number.
Prove that if $f$ is bounded above and $d = \sup{f}$, then $\lim_{x \to b^{-}}f(x) = d$
I've written a solution to this, partially based on what I've seen before so it is regurgitated:
Solution:
Let $\epsilon > 0$. We are given that $d = \sup{f}$ as such by definition $d - \epsilon$ is not a supremum. This means there exists some $c < b$ such that $d - \epsilon < f(c)$. Let $\delta = b-c$. Therefore if $x < b$ and $b-x < \delta$ then we have $c < x < b$ and by monotonicity we have $d-\epsilon < f(c) < f(x) < d$. So by definition the limit is proven to exist.
My questions come about in trying to reason out the choice of $\delta = b - c$ and how are we able to introduce the conditions: "if $x < b$" and "$b - x < \delta$"? I understand why this is a good choice for $\delta$, but I can't put the "thought process" I should be having to get it. The last few questions I've had on here revolve around this idea. I'm working on structuring my thinking instead of just throwing things against the wall.