Recently I asked this question, regarding how to deal with a quotient on a proof of the chain rule. I've come to wonder if a step in Spipvak's proof in Calculus on Manifolds needs a similar justification. The question is at the very end of the post.
Considering that Problem 1-10 requires us to prove that any linear transformation $\mathbb{R}^n\to\mathbb{R}^m$ is bounded/continuous, my guess is that the intented justification of $(6)$ tending to zero is:
$$\frac{\mu(\phi(x))}{x-a} = \frac{\mu(\phi(x))}{\phi(x)}\frac{\phi(x)}{x-a}$$
where
$$\frac{|\mu(\phi(x))|}{|\phi(x)|}\le M \ \ \ \ \text{and} \ \ \ \ \lim_{x\to a}\frac{|\phi(x)|}{|x-a|} = 0$$
for some $M\in \mathbb{R}$. So that ${\mu(\phi(x))}/(x-a)\to 0$ whenever $x\to a$. However
$$\textbf{what justifies dividing by } \phi(x)?$$
No, why bother dividing? We have for any $x\neq a$, \begin{align} \frac{\|\mu(\phi(x))\|}{\|x-a\|}&\leq \frac{\|\mu\|\cdot \|\phi(x)\|}{\|x-a\|}=\|\mu\|\cdot \frac{\|\phi(x)\|}{\|x-a\|}. \end{align} The RHS vanishes as $x\to a$ because of $(4)$. So, squeeze theorem tells you the LHS also vanishes as $x\to a$. This proves $(7)$.
Here, I wrote $\|\mu\|$ to mean the operator norm of the linear map $\mu$. But really all we care about is there exists a number $M\geq 0$ such that for all $\xi\in\Bbb{R}^m$, $\|\mu(\xi)\|\leq M\|\xi\|$.