Suppose
$\mu>0$
$K>0$
$f(x)\sim -Ke^{-\mu x}$ as $x\to\infty$
- $g(y)\sim -Ke^{-\mu y}$ as $y\to\infty$
- x is always smaller or equal y, i.e. $x\leq y$
Does this imply that $$ f(x) + g(y) \sim -Ke^{-\mu x}-Ke^{-\mu y} $$ as $x\to\infty$ or $y\to\infty$?
I think I have to show that $$ \lim_{x\to\infty}\frac{f(x)+g(y)}{-K(e^{-\mu x}+e^{-\mu y})}=1. $$
My first observation is that $g(y)=o(e^{-\mu x})$, as $x\to\infty$, isn't it?
Hence what I get is that, for $x$ large, $$ \frac{f(x)+g(y)}{-K(e^{-\mu x}+e^{-\mu y})}\leqslant \frac{f(x)+g(y)}{-2Ke^{-\mu x}}\leqslant \frac{f(x)+g(y)}{-Ke^{-\mu x}}=\frac{f(x)}{-Ke^{-\mu x}}+\frac{g(y)}{-Ke^{-\mu x}}\to 1 $$
Does this help?
Edit
Here is another approach I tried.
I would like to show that $$ \left\lvert\frac{f(x)+g(y)}{-K(e^{-\mu x}+e^{-\mu y})}-1\right\rvert\to 0\text{ as }x,y\to\infty. $$ as $x,y\to\infty$.
$$ \begin{align*} \left\lvert\frac{f(x)+g(y)}{-K(e^{-\mu x}+e^{-\mu y})}-1\right\rvert&=\left\lvert\frac{f(x)+Ke^{-\mu x}+g(y)+Ke^{-\mu y}}{K(e^{-\mu x}+e^{-\mu y})}\right\rvert\\ &\leq \frac{\lvert f(x)+Ke^{-\mu x}\rvert}{Ke^{-\mu x}}+\frac{\lvert g(y)+Ke^{-\mu y}\rvert}{Ke^{-\mu y}}\\ &=\frac{\lvert Ke^{-\mu x}(\frac{f(x)}{Ke^{-\mu x}}+1)\rvert}{Ke^{-\mu x}}+\frac{\lvert Ke^{-\mu y}(\frac{g(y)}{Ke^{-\mu y}}+1)\rvert}{Ke^{-\mu y}}\\ &=\left\lvert\frac{f(x)}{Ke^{-\mu x}}+1\right\rvert+\left\lvert\frac{g(y)}{Ke^{-\mu y}}+1\right\rvert\\ &=\left\lvert\frac{f(x)}{-Ke^{-\mu x}}-1\right\rvert+\left\lvert\frac{g(y)}{-Ke^{-\mu y}}-1\right\rvert\to 0+0=0\textrm{ as }x,y\to\infty \end{align*} $$
Does this make sense?