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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?

Rhjg
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    Technically you need to be careful about how you define $\sim$ when you take a limit in one variable, as you need to specify the asymptotic behavior of the other variable as the limit is taking place. For example, if $y$ is fixed then $f(x)+g(y) \sim g(y)$ as $x \to \infty$. In your situation simply requiring $x \leq y$ is not enough to determine whether the $g(y)$ term is significant in the $x \to \infty$ or $y \to \infty$ limits. – Ian Dec 20 '19 at 17:41
  • But in my situation, I know that $x$ is always smaller than $y$. It comes from the context of my question. – Rhjg Dec 20 '19 at 17:43
  • See my last sentence though. – Ian Dec 20 '19 at 17:46
  • What if instead of considering the limit in one variable, I consider the limit in both variables, i.e. is the $\sim$-statement true if $x\to\infty$ and $y\to\infty$? – Rhjg Dec 20 '19 at 18:09
  • Maybe I sould add another idea. As I said, I now consider $x,y\to\infty$ and I know that $x\leqslant y$. If I am not mistkaken, $g(y)\sim -Ke^{-\mu y}$ means that $g(y)=-Ke^{-\mu y} + o(-Ke^{-\mu y})$ as $y\to\infty$. But if also $x\to\infty$, this should mean that $g(y)=o(-Ke^{-\mu x})$. On the other hand, $f(x)\sim -Ke^{-\mu x}$ as $x\to\infty$ means that $f(x)=-Ke^{-\mu x}+o(-Ke^{-\mu x})$ as $x\to\infty$. But then, doesn't that mean that $A+B=-Ke^{-\mu x}+o(-Ke^{-\mu x})$ and the right-hand side is a collection of function which in particular includes $-Ke^{-\mu x}-Ke^{-\mu y}$? – Rhjg Dec 20 '19 at 18:38
  • It's just a bit weird notationally because you need to be careful of what that $o()$ actually means, since it takes a function of just $x$ as an argument but you want to say that a function of $x$ and $y$ is in that class, which is only true when the two-variable limit is taken in a suitable way. It holds when interpreted the right way, though, as you say. – Ian Dec 20 '19 at 19:35
  • I added an edit where I tried to use the equivalent definition of $\sim$ which should work here since the denominators nerver become 0. - Is that correct? – Rhjg Dec 20 '19 at 20:57

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