If $\lim_{x \to b^{-}}f(x) = \infty$ then the image of $f$ is the ray $[f(a),\infty)$ - Proof feedback

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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 $\lim_{x \to b^{-}}f(x) = \infty$ then the image of $f$ is the ray $[f(a),\infty)$

This question is a follow up to another one I did that most likely applies: If $\lim_{x \to b^{-}} f(x) = d$ then the image of $f$ is the half closed interval $[f(a),d)$ - Proof feedback

So in that question we established that the image of $f$ is the half closed interval $[f(a),d)$.

For this question, I wanted to apply this result and let $d = \infty$. Then the result will follow. I think this is correct, but something is telling me that I can't treat $\infty$'s in such a cavalier way. So do I need to be more confident in my solutions or am I right in wanting to tread more carefully due to the $\infty$ ?

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Claim 1. $\operatorname{im}f \subset [f(a), \infty)$.
Proof. Follows from monotonicity.


Claim 2. $\operatorname{im}f \supset [f(a), \infty)$.

Proof. Let $y_0 \in [f(a), \infty)$. We need to show that there exists $x_0 \in [a, b)$ such that $f(x_0) = y_0$.

Let $M = y_0$ (as in the notation in the comments).
Let $\delta$ be as in the comments. Choose any $x \in [a, b)$ that satisfies $0 < |x - b| < \delta$. Then, we have $f(x) > M$.

On the other hand, we have $f(a) \le y_0 < f(x)$.
By IVT, we see that there exists $x_0 \in [a, x)$ such that $f(x_0) = y_0$, as desired.


Equality of $\operatorname{im} f$ and $[f(a), \infty)$ now follow.