How to prove that $log(f(x)) \sim log(g(x))$

Question

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

Answer 1

• 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.