Why sum of two little "o" notation is equal little "o" notation from sum?

Question

Why sum of two little "o" notation is equal little "o" notation from sum?

$o( f(n) ) + o( g(n) ) = o( f(n) + g(n) ) ?$

For example:

• $f(n) = n^3$
• $g(n) = 1/n$

so

• $o(f(n)) = n^2$
• $o(g(n)) = 1/n^2$

and

• $o( f(n) ) + o( g(n) ) = n^2 + 1/n^2$
• $o( f(n) + g(n) ) = n^2$

Of course, I could write it like

• $o( f(n) ) + o( g(n) ) = n^2 + o( g(n) )$
• $o( f(n) + g(n) ) = n^2 + o( g(n) )$

My question is why? I don't understand it, because in first we always get two parameters.

Answer 1

In this notation we always suppose that the function appeared in the parenthesis is positive, for a counter-example of this equality when this assumption is not applied, we can take $1=o(n^2)$ and $0=o(1-n^2)$ which contradicts with $1=o(1)$.

With this observations we have $$ \left|\lim_{n\to \infty} \frac{o(f(n))+o(g(n))}{f(n)+g(n)}\right|\le \left|\lim_{n\to \infty} \frac{o(f(n))}{f(n)}+\lim_{n\to \infty} \frac{o(g(n))}{g(n)}\right|=0 $$ As we want. $\square$