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

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

• I find it confusing to have $\lim_{x\to c}$ followed by an expression that does not contain $x$ at all. Surely you are not taking the limit of a constant? – Marc van Leeuwen Dec 27 '15 at 19:25
• I'm sorry, with $f$ and $g$ I mean $f(x)$ and $g(x)$ – Gianolepo Dec 27 '15 at 19:27

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 $x\to c$.

• That depends on what the OP means by "$g(x)\ne 0$ near a point $c$". – TonyK Dec 27 '15 at 18:55
• @TonyK No problem, OP doesn't require the functions to be continuous. Hence we can (arbitrarily) define $f(0)=g(0)=1$. – Wojowu Dec 27 '15 at 19:00
• @TonyK : I would take it to mean that there is some open interval about $c$ for which for all $x$ in that interval except $x=c$ we have $g(x)\ne 0$. ${}\qquad{}$ – Michael Hardy Dec 27 '15 at 19:01
• @Wojowu : I presume you mean $f(c)$ and $g(c)$ rather than $f(0)$ and $g(0)$. But why define it at that point at all? It's just a distracting purposeless complication. ${}\qquad{}$ – Michael Hardy Dec 27 '15 at 19:15
• I'm saying "at $0$" instead of "at $c$" because your example has $x\rightarrow 0$ and not $x\rightarrow c$. – Wojowu Dec 27 '15 at 19:22

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