If $f$ is not bounded from above, then $\lim_{x \to b^{-}}f(x) = \infty$ - Feedback on attempted proofs 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.
Claim: If $f$ is not bounded from above, then $\lim_{x \to b^{-}}f(x) = \infty$
Attempt (including reasoning):
We know:
i) $f$ is a continuous monotone increasing function.
ii) $f$ is not bounded. Which means by definition: If $f$ is continuous on $[a,b)$, then $f$ is not bounded above if for all $N >0$, there exists $x \in [a,b)$ such that $f(x) > N$.
What we want:
To show: $\lim_{x \to b^{-}}f(x) = \infty$, which is a symbolic way of saying that the function diverges. This means by definition: For all $M > 0$, there exists a $\delta > 0$ such that if $0< |b - x| < \delta$ then $f(x) > M$.
Proof 1: Assume towards contradiction that $\lim_{x \to b^{-}}f(x) = d$. Since $f$ is monotonic this would imply that $f(a) \leq f(x) \leq d$ this would imply that the function $f$ is bounded above. Which would be a contradiction to our original assumption.
Proof 2: This is more of an brainstorm. I wanted to try and show the result directly. But was coming up empty. What I envision happening is that I could use the definition of the function being bounded, but where I stumble is trying to find a $\delta$ in order for me to be able to connect the bounded function definition to the definition of divergence.
Could I get feedback on Proof 1 and whether the idea for proof 2 is possible ?
Just as an advisory. I haven't been introduced to sequences formally in the textbook I'm using. I'm aware there are sequential ways of possibly doing this, but was trying to avoid them.
 A: *

*It would make more sense to not introduce the number $d$ until you actually use it, since $d$ isn't an arbitrary number here, you want to say later on "suppose $d = \lim_{x \to b^-} f(x)$".

*In Proof 1, in my opinion "Assume towards contradiction that $\lim_{x \to b^-} f(x) = d$" needs more justification. Generally, the converse of "$f$ tends to infinity" is not "$f$ tends to some finite limit".

Have you noticed that you don't really use continuity anywhere? Indeed this claim holds if $f$ is not continuous. Really the fact that $f$ is not bounded above is almost identical to the fact that $f \to \infty$, as $f$ is monotone increasing (it didn't even need to be strictly). If you want $f$ to be bigger than some given $M$, the fact that $M$ is not an upper bound gives you a $y$ such hat $f(y) > M$, and then by monotonicity, for any $x \ge y$, $f(x) > M$.
A: There is indeed a direct way of showing that the function diverges (as alluded to by your "proof 2"). For any $M>0$, you have since the function $f$ is unbounded above that $f(x)>M$ for some $x\in[a,b)$. By the fact that $f$ is monotone, any $y>x$ will have $f(y)\geq f(x)>M$. Can you infer an appropriate choice of $\delta$ from this?
