5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Marc van Leeuwen Dec 27 '15 at 19:25
  • $\begingroup$ I'm sorry, with $f$ and $g$ I mean $f(x)$ and $g(x)$ $\endgroup$ – Gianolepo Dec 27 '15 at 19:27
5
$\begingroup$

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 $x\to c$.

$\endgroup$
  • 1
    $\begingroup$ That depends on what the OP means by "$g(x)\ne 0$ near a point $c$". $\endgroup$ – TonyK Dec 27 '15 at 18:55
  • $\begingroup$ @TonyK No problem, OP doesn't require the functions to be continuous. Hence we can (arbitrarily) define $f(0)=g(0)=1$. $\endgroup$ – Wojowu Dec 27 '15 at 19:00
  • $\begingroup$ @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 $g(x)\ne 0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 27 '15 at 19:01
  • $\begingroup$ @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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 27 '15 at 19:15
  • 1
    $\begingroup$ I'm saying "at $0$" instead of "at $c$" because your example has $x\rightarrow 0$ and not $x\rightarrow c$. $\endgroup$ – Wojowu Dec 27 '15 at 19:22
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.