To prove $e^{x + o (x)} = 1 + x$ as $x \rightarrow 0$, I can do it directly:
$\lim_{x \rightarrow 0} \frac{\log (1 + x) - x}{x} \overset{\text{l'hopital}}{=}\lim_{x \rightarrow 0} \frac{(1 + x)^{- 1} - 1}{1} = 0$
But this also looks like Taylor's theorem, here's another "proof" that doesn't feel right:
By Taylor's theorem, $e^{x + o (x)} = 1 + (x + o (x)) + o (x + o (x))$. Clearly, $0 \in o (x) + o (x + o (x))$.
Is there some way to justify this?
On
It seems like you want to justify that $f(x) = o(x + o(x)) \implies f(x) = o(x)$, as $x \to 0$.
By definition
$\lim_{x \to 0} \frac{o(x)}{x} = 0 \land \lim_{x \to 0} \frac{x}{x} = 1 \implies \lim_{x \to 0} \frac {o(x) + x}{x} = 1$
By hypothesis $$\lim_{x \to 0} \frac{f(x)}{x + o(x)} = 0 \implies \lim_{x \to 0} \frac{f(x)}{x + o(x)} \cdot 1 = 0 \implies \\ \lim_{x \to 0} \frac{f(x)}{x + o(x)} \cdot \lim_{x \to 0} \frac {o(x) + x}{x} = \lim_{x \to 0} \frac{f(x)}{o(x) + x} \cdot \frac {o(x) + x}{x} = 0 \implies \lim_{x \to 0} \frac {f(x)}{x} = 0 $$
By definition, that means that $f(x) = o(x)$ as $x \to 0$.
Yes, there is. $o(x)$ is just some function $f(x)$ which satisfies $\lim_{x \to 0}\frac{f(x)}{x} = 0$. So $e^x = 1 + x + f_1(x)$ where $\lim_{x \to 0}\frac{f_1(x)}{x} = 0$. Now if $\lim_{x \to 0} \frac{f_2(x)}{x} = 0$ then $e^{x + f_2(x)} = 1 + x + f_2(x) + f_1(x + f_2(x))$. We have $$ \lim_{x \to 0} \frac{f_2(x) + f_1(x + f_2(x))}{x} = \lim_{x \to 0} \frac{f_2(x)}{x} + \frac{f_1(x + f_2(x))}{x + f_2(x)}\frac{x + f_2(x)}{x} = 0 $$ So $f_3(x) = f_2(x) + f_1(x + f_2(x)) \in o(x)$. In other words $e^{x + f_2(x)} = 1 + x + f_3(x)$ and $f_3(x) \in o(x)$ which is another way of saying $e^{x + f_2(x)} = 1 + x + o(x)$.