Considering two functions $f(x)$ and $g(x)$ and a point $x_0 \in \mathbb{R} \cup \big\{+\infty,-\infty\big\}$ we say that

$f(x)\sim g(x)$ as $x\to x_0$ $\iff$ $lim_{x\to x_0} \frac{f(x)}{g(x)}=1$

How can I prove that, if:

  1. $f(x)\sim g(x)$ as $x\to x_0$
  2. $f(x) \to l$, with $l\geq 0 \wedge l\neq 1$


$log(f(x))\sim log(g(x))$ as $x\to x_0$


I was thinking about proving that $log (\frac{f(x)}{g(x)}) =log(f(x))-log(g(x)) \to 0$ as $x\to x_0$ in order to conclude that $log(f(x))\sim log(g(x))$ as $x\to x_0$

But I don't know how to do it, can anyone help me?

Thanks a lot in advice

  • $\begingroup$ What does $\sim$ mean? That the limits are equal? $\endgroup$
    – Jimmy R.
    Jan 1 '16 at 16:16
  • $\begingroup$ This is Landau notation: $f\sim g$ means $f-g=o(g)$, or (if $g$ does not cancel) $\frac{f}{g}\to 1$. $\endgroup$
    – Clement C.
    Jan 1 '16 at 16:18
  • Case $\ell >0$:

You have $g(x) = f(x) + o(f(x)) = \ell +o(1)$ around $x_0$, by applying both hypotheses. Then, $\ln f(x) \xrightarrow[x\to x_0]{} \ln \ell\neq 0$, i.e. $\ln f(x) = \ln \ell+o(1)$; and $\ln g(x) = \ln (\ell +o(1)) = \ln \ell + \ln(1 +o(1)) = \ln \ell+o(1)$.

This directly implies $\ln f(x) \sim_{x\to x_0} \ln g(x)$, are both are equivalent to $\ln \ell$.

  • Case $\ell = 0$ (actually also subsums the one above) $\ln g(x) - \ln f(x) = \ln \frac{g(x)}{f(x)} \xrightarrow[x\to x_0]{} \ln 1 = 0$ (using that $\frac{g}{f} \to 1$ as $f\sim g$, and continuity of the logarithm), and since $\ln f(x) \to \ln\ell \neq 0$ we do get $\ln f(x) - \ln g(x) = o(\ln f(x))$, giving the equivalence.
  • $\begingroup$ Very nice! If I may ask, in the case that $l=+\infty$ is it correct to deduce from $ln(f(x))=ln l+o(1)=ln(g(x))$ that $ln(f(x))\sim ln(g(x))$, i.e. $ln(f(x))=ln(g(x))+o(ln(g(x)))$? Since in general it is not because $x^2\to \infty$, $x\to \infty$ but $x^2 \not \sim x$ as $x\to +\infty$. $\endgroup$
    – Gianolepo
    Jan 1 '16 at 16:56
  • 1
    $\begingroup$ The argument for the first case does not work for $\ell = \infty$, only for finite $\ell$ (use the second one for that). You cannot write $\ln \ell + o(1)$ when $\ell=\infty$. $\endgroup$
    – Clement C.
    Jan 1 '16 at 16:58
  • $\begingroup$ Thanks! And in the very first equality I'm totally ok with $g(x)=f(x)+o(f(x))$ but to get to $g(x)=l+o(1)+o(f(x))=l+o(1)$ did you "substitute" $f(x)=l+o(1)$ inside the little o? I read on my textbook that in general from $f(x) \sim g(x)$ we cannot deduce that $f(x)$ and $g(x)$ have the same limit (that could also not exist separately) but is in this case possible because from the hypotesis, $f(x)$ has a limit, equal to $l$ and automatically we get that $g(x)$ must have the same limit? $\endgroup$
    – Gianolepo
    Jan 1 '16 at 17:15
  • 1
    $\begingroup$ Yes: you can see it using the ratio $f/g\to 1$ if that helps, but basically if $f\sim g$ and $f\to \ell$, then $g\to \ell$ must hold. $\endgroup$
    – Clement C.
    Jan 1 '16 at 17:18

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.