Example
37
[1FD]We informally state this property.
If \(n≥ m≥ 1\) then \(o(x^ n)+o(x^ m)=o(x^ m)\).
To prove it, we convert it into a precise statement. First of all, let’s rewrite it like this.
If \(f(x)=o(x^ n)\) and \(g(x)=o(x^ m)\) then \(f(x)+g(x)=o(x^ m)\).
So let’s prove it. From the hypotheses,
\[ \lim _{x→ 0}f(x)x^{-n}=0 ~ \text{and}~ ~ \lim _{x→ 0}g(x)x^{-m}=0\quad \]
then
\[ \lim _{x→ 0}\frac{f(x)+g(x)}{x^{m}}= \lim _{x→ 0}\frac{f(x)}{x^{m}}+ \lim _{x→ 0}\frac{g(x)}{x^{m}}= \lim _{x→ 0} x^{n-m} \frac{f(x)}{x^{n}}+0=0. \]