Using logarithms to simplify fractional limits to infinity

Question

$$\lim_{x \to \infty} \frac{\ln f(x)}{\ln g(x)} = +\infty \Rightarrow \lim_{x \to \infty} \frac{f(x)}{g(x)} = +\infty$$

Is this true? If it is, does anyone have any tips on how to prove it? I feel like it's a simple matter of the nature of the logarithmic function, but I'm just not sure how to set up the argument.

Edit: I missed some edge cases in my original question. In order to avoid them, I also know that as $x \to \infty$, $f(x), g(x) \to +\infty$. Given this, is the statement true?

Answer 1

It is not true. If $g(x) \to 1, \ln (g(x)) \to 0$ which can make the first fraction diverge even if $f(x)$ is constant. Then the ratio without the logs goes to the value of $f(x)$.

For example, let $g(x)=1+\frac 1x, f(x)=3$. Then $\frac {\ln f(x)}{\ln g(x)}=\frac {\ln 3}{\ln(1+\frac 1x)}\approx x\ln(3)\to +\infty$, but $\frac {f(x)}{g(x)}=\frac 3{1+\frac 1x}\to 3$

Added: The only problem comes when $g(x) \to 1$. If you can show $g(x) \gt k \gt 1$ you have $\frac 1{\ln (g(x))} \gt \frac 1{\ln (k)}$ so the fact that the limit of the logs is infinite means the limit of the log of $f(x)$ is infinite. The ratio of the functions then has to go to infinity as well.

Answer 2

If $\lim_{x \to \infty} \log(|f|)/\log(|g|) = \infty,$ then for each $K>0$, there is some $X>0$ s.t. whenever $|x|>X$, $$ |\log(|f|)| > K |\log(|g|)| \implies \log(|f|) > K|\log(|f|)| \text { or } \log(|f|) < -K |\log(|g|)|. $$ Take exponential: $$ |f| > \mathrm e^K \exp(|\log (|g|)|) $$ or $$ |f| <\mathrm e^{-K}\exp(|\log (|g|)|). $$ If $|g|>1$ for sufficiently large $x$, then $$ |f| > \mathrm e^K |g| \text{ or } |f| < \mathrm e^{-K} |g|. $$ According to these, you could take $f(x) =\mathrm e^{-x}$, $g(x) =\mathrm e$ to satisfy the second case above, then $\log(f)/\log(g) = -x \to \infty$, but $f/g \to 0$.

P.S. The meaning of $\infty$ is not specified, so I take the most general definition.

Answer 3

$\lim_{x \to \infty} \frac{\ln f(x)}{\ln g(x)} = +\infty \Rightarrow \lim_{x \to \infty} \frac{f(x)}{g(x)} = +\infty $

You need the additional assumption that $g(x) \to \infty$.

If $\lim_{x \to \infty} \frac{\ln f(x)}{\ln g(x)} = +\infty $ then, for any $c > 0$, there is an $x(c)$ such that $\frac{\ln f(x)}{\ln g(x)} \gt c$ for $x > x(c)$.

Then $\ln (f(x)) \gt c\ln (g(x)) $ so, if $c > 1$ and $g(x) \to \infty$,

$\begin{array}\\ \ln(f(x)/g(x)) &=\ln(f(x))-\ln(g(x))\\ &>c\ln(g(x))-\ln(g(x))\\ &=(c-1)\ln(g(x))\\ &\to \infty\\ \end{array} $

so $f(x)/g(x) \to \infty$.