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Let $f,g,h,j$ be functions of $x$.

Suppose $f\sim g$ as, say $x\to\infty$ and $h\sim j$ as $x\to\infty$ which, by definition, means that $$ f-g=\mathcal{o}(g)\text{ and }h-j=\mathcal{o}(j)\textrm{ as }x\to\infty. $$

Does this imply that $f +h\sim g+j$ as $x\to\infty$.

I would think so. It is to show that $$ f+h - (g+j)=\mathcal{o}(g+j), x\to\infty. $$ So let $\varepsilon >0$. Then $$ \lvert f+h-(g+j)\rvert=\lvert f-g+h-j\rvert\leqslant\lvert f-g\rvert+\lvert h-j\rvert\leqslant \varepsilon g+\varepsilon j=\varepsilon (g+j) $$ for all $x>X$ for some $X:=\max\{X_1,X_2\}$, where $$ \lvert f-g\rvert\leqslant \varepsilon g~\forall x>X_1 $$ and $$ \lvert h-j\rvert\leqslant\varepsilon j~\forall x>X_2 $$ where $X_1,X_2$ exist by assumption.

But this should mean exactly that $f+g\sim h+j$ as $x\to\infty$.

However, I am not sure. Maybe there is some counterexample as it usually is the case when I try to prove things like this.

Rhjg
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    If you have positivity assumptions, it works. Without that, let $f = g \colon x \mapsto x$, $h \colon x \mapsto -x + 1$, $j \colon -x + 1/x$ for a counterexample. – Daniel Fischer Dec 09 '19 at 20:12
  • What do you mean with positivity assumptions? In your example, I get that $\lvert f+h-(g+j)\rvert=1-\frac{1}{x}$ and $g+j=\frac{1}{x}$. Hence the first expression cannot be smaller than 1. – Rhjg Dec 09 '19 at 20:19
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    By "positivity assumptions" I mean an assumption/condition that $f,g,h,j$ are positive functions. If they are, then no cancellation like in my example can occur, and you have $f + h \sim g + j$. If the signs of the functions can be different, cancellation may occur and then you need not have asymptotic equality. – Daniel Fischer Dec 09 '19 at 20:23
  • Thanks, got it. – Rhjg Dec 09 '19 at 20:24

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