$\lim_{x \to c} f-g=0\iff f \sim g$? 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 $f(x)=x$ and $g(x)=4x$ then $\lim\limits_{x\to0} (f(x)-g(x))=0$ and $\lim\limits_{x\to0} \dfrac{f(x)-g(x)}{g(x)} = -3 \ne 1$.
$\lim\limits_{x \to c} f-g=0 \implies \lim_{x\to c} \dfrac{f-g}{g}=0$ does not hold if $g(x)\to 0$ as $x\to c$.
