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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?

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1 Answer 1

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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 $|\frac{f_1(x)* f_2(x)}{g_1(x) * g_2(x)}| = |\frac{f_1(x)}{g_1(x)}| * |\frac{f_2(x)}{g_2(x)}|$.

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  • $\begingroup$ Ok, the second one is clear now..of course. But then subadditivity for the limsup also holds for $\limsup(a+b) \leq \limsup(c) + \limsup(d)$ where $(a + b) \leq (c +d)$ ? $\endgroup$
    – TestGuest
    Mar 20, 2013 at 14:50
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    $\begingroup$ Yes because $\limsup(a + b) \leq \limsup(c + d) \leq \limsup(c) + \limsup(d)$. $\endgroup$
    – ferson2020
    Mar 20, 2013 at 14:54

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