By definition two functions $f$ and $g$ are asymptotically equivalent near a point $a$ if and only if there is a function $u$, which has a limit of $1$ at $a$ and satisfies: $f(x) = u(x) g(x)$.
Out of the many properties of these functions, addition isn't one of them, however according to this it is, but only when $g_1$ and $g_2$ are of the same sign and never both equal zero for a set $x$ near $a$. Of course with $f_1$ equivelant to $g_1$ and $f_2$ equivalent to $g_2$.
I had a go at proving this using the equation in the definition $f = gu$, I reached:
$$f_1+f_2 = \frac{u_1 g_1 + u_2 g_2}{g_1+g_2} (g_1 + g_2).$$
However I got stuck at proving that the limit of the portion
$$\frac{u_1 g_1 + u_2 g_2}{g_1+g_2}$$
tends to $1$ when $x$ nears $a$.
Any idea on how to do it? or am I trying to prove a false property?
Best Answer
Let us assume that $g_1$ and $g_2$ are both positive for $x$ sufficiently close to $a$ (the negative case is similar). Fix $\epsilon>0$; then for $x$ sufficiently close to $a$, $|u_1-1|$ and $|u_2-1|$ are both smaller than $\epsilon$. We then have $$\left|\frac{u_1 g_1 + u_2 g_2}{g_1+g_2}-1\right|=\left|\frac{(u_1-1)g_1+(u_2-1)g_2}{g_1+g_2}\right|\leq\frac{|u_1-1|g_1+|u_2-1|g_2}{g_1+g_2}< \frac{\epsilon g_1+\epsilon g_2}{g_1+g_2}=\epsilon$$ when $x$ is sufficiently close to $a$. Since $\epsilon>0$ is arbitrary, this proves $f_1+f_2$ is asymptotically equivalent to $g_1+g_2$.