If $\displaystyle f_1(x)\sim f_2(x) $ and $\displaystyle g_1(x)\sim g_2(x) $, is $\displaystyle f_1(x)+g_1(x)\sim f_2(x)+g_2(x) $?
It is.
We have to show that $\displaystyle \dfrac{f_{1}(x)+g_{1}(x)}{f_{2}(x)+g_{2}(x)}\sim1$.
and we know that $\displaystyle \dfrac{f_{1}(x)}{f_{2}(x)}\sim1$ and $\displaystyle \dfrac{g_{1}(x)}{g_{2}(x)}\sim1$.
Therefore, we have
$\displaystyle \dfrac{f_{1}(x)+g_{1}(x)}{f_{2}(x)+g_{2}(x)}=\dfra c{1+g_{1}(x)/f_{1}(x)}{\big(f_{2}(x)/f_{1}(x)\big)+\big(g_{2}(x)/f_{1}(x)\big)}$
........................$\displaystyle \sim\dfrac{1+\big(g_{1}(x)/f_{1}(x)\big)}{1+\big(g_{2}(x)/f_{1}(x)\big)}$
........................$\displaystyle =\dfrac{\big(f_{1}(x)/g_{1}(x)\big)+1}{\big(f_{1}(x)/g_{1}(x)\big)+\big(g_{2}(x)/g_{1}(x)\big)}$
........................$\displaystyle \sim\dfrac{\big(f_{1}(x)/g_{1}(x)\big)+1}{\big(f_{1}(x)/g_{1}(x)\big)+1}\equiv1$.