Regarding little oh as a real number

Question

In Strichartz's $\textit{The Way of Analysis}$ he justifies differentiability of $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ implies continuity by taking the limit as $x \rightarrow y$ of $$f(x) = f(y)+df(y)(x-y)+o(|x-y|)$$ Obviously $f(y)$ is a constant and I was able to justify to myself that $df(y)(x-y) \rightarrow 0$ as $x \rightarrow y$, but $o(|x-y|) \rightarrow 0$ seems absurd because as far as I am concerned little oh doesnt represent a real number (as a note I am familiar with the limit definition). He goes on to further show that $f$ is also Lipschitz continuous by writing $$|f(x)-f(y)| = |df(y)(x-y)+o(|x-y|)| \le |df(y)(x-y)|+|o(|x-y|)|$$ (again regarding little oh as a real number) and then noticing $|o(|x-y|)| \le |x-y|$ and $|Ax| \le M|x|$ for any matrix $A$. Can someone help me justify why we can regard little oh as a real number and treat it algebraically?

Answer 1

$o(|x-y|)$ is a placeholder for the remainder term $$r(x,y) := f(x) - f(y) - df(y)(x-y)$$ and indicates that $\frac{|r(x,y)|}{|x-y|} \to 0$ as $x \to y$.