Is it true that, for two infinite functions, $\lim_{x->x_0}\frac{f(x)}{g(x)}=1$ entails $\lim_{x->x_0}[f(x)-g(x)]=0$?

Question

Let $f,g$ be real functions defined in a neighborhood of $x_0\in \mathbb{R}$ such that $g(x)>0$,$\lim_{x\rightarrow x_0}f(x)= \lim_{x\rightarrow x_0}g(x)=+\infty$, and $\lim_{x\rightarrow x_0}\frac{f(x)}{g(x)}=1$. Is it true that $$\lim_{x\rightarrow x_0}[f(x)-g(x)]=0?$$

Answer 1

The statement is false: consider this counterexample, for $x_0=0$. $$ f(x)=\frac1{x^2} ; g(x)=\frac1{x^2} -1$$ It is easy to see that $\lim_{x\to 0}f(x)=\lim_{x\to 0}g(x)=+\infty $, but $f(x)-g(x)=1$.

Answer 2

$$f(x)-g(x)=(\frac{f(x)}{g(x)}-1)g(x) \to(1-1)*0=0$$

for $x \to x_0$

Answer 3

If $\lim \limits_{x \to x_0} f(x) = \lim \limits_{x \to x_0} g(x) = 0$, then $\lim \limits_{x \to x_0} \{f(x) - g(x)\} = \lim \limits_{x \to x_0} f(x) - \lim \limits_{x \to x_0} g(x) = 0 - 0 = 0$. You don't need any kind of assumption on the ratio of $f$ and $g$.

Edit: This was the answer to the old question. The new statement is incorrect. Consider e.g. $f(x) = \frac{1}{x^2}$ and $g(x) = \frac{1}{x^2} + \frac{1}{|x|}$ around $x_0 = 0$.