Is this a valid proof of $o(x^n) + o(x^m) = o(x^{\min(n,m)}) $? I made this proof about the little o algebra , My aim is to prove :
$$o(x^n) + o(x^m) = o(x^{\min(n,m)}) \:\:\:\:\:\:\:x \to 0 $$
So , if $f(x) = o(x^n)$ and $g(x) = o(x^m)$ then without loss of generality take $\min(n,m) = n $ then :
$$\lim_{x \to 0}{\dfrac{f(x)+g(x)}{x^n}} = \lim_{x \to 0}{\dfrac{f(x)}{x^n}} + \lim_{x \to 0}{\dfrac{g(x)}{x^n}} = 0  + \lim_{x \to 0}{\dfrac{g(x)}{x^n}\cdot\dfrac{x^{m-n}}{x^{m-n}}} = 0 + \lim_{x \to 0}{\dfrac{g(x)}{x^m}\cdot\lim_{x \to 0}{x^{m-n}}} = 0 $$
then   $o(x^n) + o(x^m) = o(x^{\min(n,m)}) \:\:\:\:\:\:\:x \to 0$
I'm a bit confused about the little o notation so I'm not sure if my reasoning is valid.
EDIT :
New way of doing it as suggested in the comments :
if $n<m$ then $x^m = o(x^n)$ so I can write :
$$o(x^n) + o(x^m) = o(x^n) + o(o(x^n)) = 
o(x^n) + o(x^n) = o(x^n) \:\:\:\:\:\:\:x \to 0$$
 A: Go back to the definition : 
A function $w$ is said to be in the class $o(h(x))$ while $x \to a$ with $h$ a non-negative function if there exists a non-negative function $\varepsilon$ with $\lim_{x\to a} \varepsilon (x)=0$ such that $|w(x)| \leqslant \varepsilon(x) h(x)$ in a neighboorhood of $a$. 
To prove the statement take $f$ of class $o(x^n)$ and $g$ of class $o(x^m)$ at $0$, and suppose w.l.o.g that $n \leqslant m$. There exists $\varepsilon_f$ and  $\varepsilon_g$ with limit $0$ at $0$ such that
\begin{align}
|f(x)| &\leqslant \varepsilon_f(x) x^n \\
|g(x)| &\leqslant \varepsilon_g(x) x^m
\end{align}
in a neighborhood of $0$. 
Notice that $\varepsilon_g(x)x^{m-n}$ still has limit $0$ at $0$. Hence
\begin{align}
|f(x) + g(x)| \leqslant |f(x)|+|g(x)| \leqslant \underbrace{\left( \varepsilon_f(x) + \varepsilon_g(x)x^{m-n}\right)}_{\varepsilon(x)}x^n
\end{align}
with $\varepsilon(x) \to 0$ at $0$. Consequently, $f+g$ is of class $o(x^n)$
Remark : it looks the same as you did. The main difference is there is no division. If you're trying to prove $w = o(h)$ with $h$ non-negative but vanishing infintly many times in a neighborhood of $a$, you cannot divide by $h$ to prove the statement, but this proof always works.
