We had to proove that if $f_1(x) = O(g_1(x))$ and $f_2(x) = O(g_2(x))$ for $g_i(x)$ > 0 then
$i) f_1(x) + f_2(x) = O(g_1(x) + g_2(x))$ and $ii) f_1(x) + f_2(x) = O(g_1(x)g_2(x))$
Now I have questions regarding parts of the proofs:
For i) it is stated that $\limsup_{x \rightarrow a} |\frac{f_1(x) + f_2(x)}{g_1(x) + g_2(x)}| \leq \limsup_{x \rightarrow a}|\frac{f_1(x)}{g_1(x) + g_2(x)}| + \limsup_{x \rightarrow a}|\frac{f_2(x)}{g_1(x) + g_2(x)}|$
my question: I know about the subadditivity of the limes superior, but also think that this presupposes that: $|\frac{f_1(x) + f_2(x)}{g_1(x) + g_2(x)}| = |\frac{f_1(x)}{g_1(x) + g_2(x)}| + |\frac{f_2(x)}{g_1(x) + g_2(x)}|$ which from the triangle inequality is not necessarily true, is it?
Regarding ii) it is stated simply that $\limsup_{x \rightarrow a} |\frac{f_1(x)* f_2(x)}{g_1(x) * g_2(x)}| = \limsup_{x \rightarrow a}|\frac{f_1(x)}{g_1(x)}|*\limsup_{x \rightarrow a}|\frac{f_2(x)}{g_2(x)}|$ - why is that?
It is generally true that $|\frac{f_1(x) + f_2(x)}{g_1(x) + g_2(x)}| \leq |\frac{f_1(x)}{g_1(x) + g_2(x)}| + |\frac{f_2(x)}{g_1(x) + g_2(x)}|$, but you do not necessarily have equality. This is enough to prove that $\limsup_{x \rightarrow a} |\frac{f_1(x) + f_2(x)}{g_1(x) + g_2(x)}| \leq \limsup_{x \rightarrow a}|\frac{f_1(x)}{g_1(x) + g_2(x)}| + \limsup_{x \rightarrow a}|\frac{f_2(x)}{g_1(x) + g_2(x)}|$.
$\limsup_{x \rightarrow a} |\frac{f_1(x)* f_2(x)}{g_1(x) * g_2(x)}| = \limsup_{x \rightarrow a}|\frac{f_1(x)}{g_1(x)}|*\limsup_{x \rightarrow a}|\frac{f_2(x)}{g_2(x)}|$ follows from the fact that $|\frac{f_1(x)* f_2(x)}{g_1(x) * g_2(x)}| = |\frac{f_1(x)}{g_1(x)}| * |\frac{f_2(x)}{g_2(x)}|$.