Little $o$ notation in the exponent How would you tackle the following expression:
$$\phi(x)^{(1+o(1))}=1$$
where $\phi:\mathbb{R}\rightarrow\mathbb{R}^+$ is continuous? $o(1)$ relates to $x\rightarrow\infty$. The specific function is known and I would like to prove the validity of the above expression for $\phi$. I am not sure how to proceed. What is a first step?
Might it be admissible to logarithmize the above Landau relation?
What is, after all, the correct interpretation for the expression $\phi(x)^{(1+o(1))}=1$?
 A: Two Interpretations
I discussed this with vitamin d in chat and mentioned that the little-o on the left-hand side of an "equation" was odd, but extrapolating from its meaning on the right, $f=o(g)$ would be more accurately be stated as
$$
f\in o(g)=\left\{h:\lim_{x\to\infty}\frac{h(x)}{g(x)}=0\right\}
$$
However, Elliot Yu commented below that in this answer it is said that when $o(g)$ appears on the left and $o(h)$ appears on the right, that $o(f)+o(g)=o(h)$ represents an inclusion of classes:
$$
o(f)+o(g)\subset o(h)
$$
that is, any element of $o(f)$ plus any element of $o(g)$ is an element of $o(h)$.

*

*However, in the case of this question, $o(1)$ appears only on one side of the "equation", so it is not clear whether this is meant to be a comparison of classes, or to say that $1\in\phi(x)^{1+o(1)}$ (which is what my answer below had assumed).


*If we do use the class comparison interpretation we would have
$$
\phi(x)^{1+o(1)}\subset\{1\}
$$
which would say that for any $u\in o(1)$,
$$
\phi(x)^{1+u(x)}=1
$$
which implies that $\phi(x)=1$ for all $x\in\mathbb{R}$.

Interpretation
The proper interpretation of
$$
\phi(x)^{1+o(1)}=1\tag1
$$
is that there exists a $u(x)$ so that $\lim\limits_{x\to\infty}u(x)=0$ and
$$
\phi(x)^{1+u(x)}=1\tag2\\[6pt]
$$

Conclusion
$\lim\limits_{x\to\infty}u(x)=0$ guarantees that there is an $x_0$ so that
$$
\begin{align}
x\gt x_0
&\implies|u(x)|\lt1\tag{3a}\\[6pt]
&\implies1+u(x)\gt0\tag{3b}
\end{align}
$$
Therefore, $(2)$ and $\text{(3b)}$ say that
$$
x\gt x_0\implies\phi(x)=1\tag4\\[6pt]
$$
Thus, if for $x$ sufficiently large, $\phi(x)=1$, your function satisfies $(1)$.
A: It is how I understand your problem, I may be wrong so let me know if it is the case :
You can rewrite your $o(1)$ as a function $\varepsilon$ with $\varepsilon(x) \rightarrow0$ when $x\rightarrow \infty$.
It implies that you can find $x_0 \in \mathbb{R}$ such that :
$$\forall x\geq x_0, |\varepsilon(x)| \leq 1/2$$
Therefore,
$$\forall x\geq x_0,  \quad 1/2 \leq 1+\varepsilon(x) \leq 3/2$$
Hence,
$$\forall x \geq x_0, \varphi(x)^{1+\varepsilon(x)} = 1$$
And as $\varphi$ is strictly positive then,
$$\forall x\geq x_0,(1+\varepsilon(x)) \ln \varphi(x) =0$$
But $\forall x \geq x_0, 1+\varepsilon(x)\geq1/2$ therefore:
$$\forall x\geq x_0, \ln \varphi(x) = 0 \quad \mathrm{i.e.} \quad \varphi(x) =1$$
To conclude, your expression for me means that for $x$ large enough, $\varphi(x) = 1$.
