$\lim_{x \to c} f-g=0\iff f \sim g$? Considered two functions $f$ and $g$ (with $g\neq 0 $ near a point $c$)
$$\lim_{x\to c} \frac{f}{g}=1 \iff f \sim g$$
I'm trying to understand if it is true that
$$\lim_{x\to c} \frac{f}{g}=1 \iff \lim_{x \to c} f-g=0$$
I tried in this way:
For the $\implies$: 
$$\lim_{x\to c} \frac{f-g}{g}=0 \implies f-g=o(g) \implies f-g=o(1)$$
For the $\Leftarrow$: 
$$\lim_{x \to c} f-g=0 \implies \lim_{x\to c} \frac{f-g}{g}=0 \implies \lim_{x\to c} \frac{f}{g}=1$$
I feel that there is something wrong with it, am I missing something?
Thanks a lot for your help

If $f(x)=(\ln |x|)^2+\ln |x|$ and $g(x)=(\ln |x|)^2$,
then  $$\lim\limits_{x\to 0} \dfrac{f(x)}{g(x)} =\lim\limits_{x\to 0} \dfrac{(\ln |x|)^2+\ln |x|}{(\ln |x|)^2}=\lim\limits_{x\to 0} \dfrac{1+1/\ln |x|}{1} = 1$$ and 
$$\lim\limits_{x\to0} (f(x)-g(x))=\lim\limits_{x\to0} (\ln |x|)=-\infty\neq0.$$
