I'm reading Spivak's calculus, but the theorem statements leave out some details. Is the theorem and proof I wrote down correct? Did I miss something? Can I weaken the hypothesis?
Theorem: Let $g$ be a real-valued function that is defined in a neighborhood $A$ of $a \in \mathbb{R}$ that does not necessarily contain $a$ and assume $\lim_{x \to a} g(x) = l \neq 0$. Then, $\lim_{x \to a} \left(\frac{1}{g}\right)(x) = \frac{1}{l}$
Proof: By hypothesis, we know that $\forall ε >0: \exists \delta > 0: \forall x \in A: (0 < |x-a| < \delta \implies |g(x) - l| < ε)$
Taking $ε = \frac{|l|}{2}$, we find that there is $\delta_1 > 0$ such that $\forall x \in A$:
$$0 < |x-a| < \delta_1 \implies |g(x)-l| < \frac{|l|}{2}$$
So, if $l > 0$, it follows that $0 <\frac{l}{2} = l - \frac{l}{2} <g(x)$ and if $l < 0$, it follows that $g(x) < \frac{l}{2} < 0$, provided that $x$ satisfies $0 < |x-a| < \delta_1$
Hence, for $x \in A$ satisfying $0 < |x-a| < \delta_1$, we find that $|g(x)| >\frac{|l|}{2}$. In particular $g(x) \neq 0.$
Now, fix $ε > 0$. Choose $\delta_2 > 0$ such that for all $x \in A$:
$$0 < |x-a| < \delta_2 \implies |g(x)-l| <ε' = \frac{|l|^2 ε}{2}$$
Hence, for $x \in A$ satisfying $0 < |x-a| < \delta := \min\{\delta_1, \delta_2\}$, we find:
$$\left|\frac{1}{g(x)} - \frac{1}{l}\right| = \frac{|g(x)-l|}{|g(x)||l|}< \frac{2ε'}{|l|^2} = ε$$
and this ends the proof.