Standard result on limits? But what is it? Book's Question
Suppose $g : \mathbb R\to \mathbb R$ is continuous at $0$, and let $h : (0, ∞) → R$ be defined by $h(x) = g(1/x)$. Prove that if $ h(n) > 0 $ for all $n \in \mathbb{N}$ then $g(0) ≥ 0$.
Book's Answer
If $g$ is continuous at $0$, then $g(0) = \lim g(1/n)$ because $1/n → 0$ as
$n → ∞$.
However, $g(1/n) = h(n) > 0 $ for every $n$, so, by standard results about limits, we must have $lim
g(1/n) ≥ 0$. Hence, $g(0) ≥ 0.$
My understanding and questions

*

*First observation: We are told g is continuous at $0$, we also know the function $1/x$ is continuous on $(0, ∞)$, note this isn't $[0, ∞)$, and the domain of $h$ is $(0, ∞)$. So to me, we cannot say h is continuous at $0$, is this correct? As $h = g(f(x))$ it is the composition of a continuous function at $0 \ g(x)$, and non-continuous function at $0 \ f(x) = 1/x$ ?


*$g(Z)$ is continuous at $0$, therefore we have some $g(Z_n)$ such that $g(Z_n) \to 0$ as $n \to ∞$. So now at this point are we supposed to arbitrarily take the sequence $Z_n$ as $1/n$? Using the kind of foreshadowing hint that $h(x) = g(1/x)$ ?


*We now take $x = n$, as we have $h(x) = g(1/x) > 0$ as us given to us. Now we use some standard results about limits...but i'm unsure what this is...i thought it might be a standard result of functions that the composition of two continuous functions is continuous, but as mentioned im not sure that is the case here...and obviously it says limits.
Thanks!
 A: 
$g(Z)$ is continuous at $0$, therefore we have some $g(Z_n)$ such that $g(Z_n) \to 0$ as $n \to \infty$. So now at this point are we supposed to arbitrarily take the sequence $Z_n$ as $1/n$?

We know that $g(Z_n)$ approaches $g(0)$, not $0$.
Rather than "we have SOME converging sequence $Z_n$ where $\lim_{n \to \infty} g(Z_n) = g\left( \lim_{n \to \infty} Z_n \right)$", the thing about a continuous function is "for EVERY converging sequence $Z_n$, we must have $\lim_{n \to \infty} g(Z_n) = g\left( \lim_{n \to \infty} Z_n \right)$". So we're allowed to choose any sequence we want; we choose $Z_n = \frac{1}{n}$ because those are the points where the problem gives us information about the values of $g$.
If that still seems like "using foreshadowing", we could word it more like:

If $n \in \mathbb{N}$, then $h(n) = g\left(\frac{1}{n}\right) > 0$. The sequence $z_n = \frac{1}{n}$ converges to zero. So since $g$ is continuous, the sequence $g(z_n)$ must converge to $g(0)$.

I'm also not sure exactly what "standard results" the author means. Maybe "if all elements in a converging sequence are in a closed set, the limit is also in that closed set"? (using here the closed set $[0,\infty)$). It kind of sounds like the author wanted to say "this is obviously true" without figuring out the actual justification.
In any case, it's not very hard to prove that statement, or more specifically that if $x_n > 0$ for all $n$ then $\lim_{n \to \infty} x_n \geq 0$, from the $N$-$\epsilon$ definition of the sequence limit.
