Asymptotic behaviour of a sum of functions.

Question

Suppose we have functions $f(x),g(x),u(x),v(x)$ such that $f(x)\sim g(x)$ and $u(x)\sim v(x)$ as $x\rightarrow x_0$. How can I show that $f(x)+u(x)\sim g(x)+v(x)$ as $x\rightarrow x_0$?

I'm using the following definition:

$$f(x)\sim g(x) \text{ as } x\rightarrow x_0\iff \underset{x\rightarrow x_0}{\lim} \frac{f(x)}{g(x)}=1.$$

I'm able to show the similar properties for the product $f(x)u(x)$ and the quotient $f(x)/u(x)$, using the product rule for limits, but I cannot show it for the sum. Could anyone help?

EDIT: Assuming $g(x)+v(x)\ne 0$ nor does it approach 0.

Answer 1

Partial answer: you need some more assumptions for this to be valid. As it stands this is false because $g+v$ may be $0$. Take $x_0=0,f(x)=x, u(x)=x^{2}-x,g(x)=x,v(x)=-x$ for example. Assuming that $g(x)+v(x)$ stays away from $0$ and that all functions are positive you can prove this as follows: $\frac {f(x)+u(x)} {g(x)+v(x)}-1=\frac {g(x)} { g(x)+v(x)} (\frac {f(x)} {g(x)}-1)+\frac {v(x)} { g(x)+v(x)} (\frac {u(x)} {v(x)}-1)$ from which the result follows easily.

Answer 2

The first claim is not ture. The following is an counterexample. Take $f(x)=-x, g(x)=-\sin x, u(x)=x, v(x)=e^x-1$. Then, we have $f(x)\sim g(x), u(x)\sim v(x)$ as $x\to 0.$ However, $$\lim\limits_{x\to 0}\frac{f(x)+g(x)}{u(x)+v(x)}=\lim\limits_{x\to 0}\frac{0}{e^x-1-\sin x}=0.$$