$\lim_{x \to c} f-g=0\iff f \sim g$?

Question

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

Answer 1

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 $x\to c$.

Answer 2

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 $$\lim\limits_{x\to0} (f(x)-g(x))=\lim\limits_{x\to0} (\ln |x|)=-\infty\neq0.$$