I was able to show that the limit of $\log(f(x))$ will have to be infinity and limit is an undetermined part of the type infinity over infinity, so I used L'Hopital's Rule, but I couldn't conclude anything else
Let $f$ and $g$ be functions of $\mathbb R$ in $\mathbb R$ and $\displaystyle\lim_{x\to+\infty} g(x)=+\infty$.
Show that if there exists $c>1$ such that $\displaystyle\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)}}=c$ then $ \displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}=+\infty$
Best Answer
This is not rigorous , it is just to show the Intuition behind why this claim must be true.
It may be modified to make it rigorous.
$\displaystyle\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)}}=\lim_{x\to+\infty} c$
$\displaystyle\lim_{x\to+\infty} {\log{f(x)}}=\lim_{x\to+\infty} c{\log{g(x)}}$
$\displaystyle\lim_{x\to+\infty} e^{\log{f(x)}}=\lim_{x\to+\infty} e^{c{\log{g(x)}}}$
$\displaystyle\lim_{x\to+\infty} {f(x)}=\lim_{x\to+\infty} {g(x)^c}$
$\displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}=\lim_{x\to+\infty} \frac{g(x)^c}{g(x)}$
$\displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}=\lim_{x\to+\infty} {g(x)^{c-1}}$
We are given :
$c \gt 1$ [[ $c-1 \gt 0$ ]]
$\displaystyle\lim_{x\to+\infty} g(x)=+\infty$
Hence :
$\displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}={+\infty}^{c-1}=+\infty$