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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 $l\geq 0 \wedge l\neq 1$

Then

$log(f(x))\sim log(g(x))$ 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

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  • $\begingroup$ 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
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  • 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.
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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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