$\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)}}=c$ then $\lim_{x\to+\infty} \frac{f(x)}{g(x)}=+\infty$ 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$
 A: 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 :
$\displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}={+\infty}^{c-1}=+\infty$
A: A stronger conclusion can be proved. For $1\le a<c$ we get $$\log{f(x)\over g(x)^{a}}= \log f(x)-a\log g(x) \\ =\log g(x)\left [{\log f(x)\over \log g(x)}-a\right ]$$ The expression in the square brackets tends to $c-a>0.$ Therefore $$\lim_{x\to \infty} \log{f(x)\over g(x)^{a}}=\infty$$ hence $$\lim_{x\to \infty} {f(x)\over g(x)^a}=\infty$$ The original conclusion is obtained with $a=1.$
A: We may think in this way: For large $x$, $\frac{\log f(x)}{\log g(x)}\approx c$.
Re-arranging terms, we have that $f(x)\approx[g(x)]^{c}$. Therefore
$\frac{f(x)}{g(x)}\approx[g(x)]^{c-1}\rightarrow+\infty$.
Now, we make everything formal and rigorous: Choose $a$ such that
$1<a<c$. Since $\lim_{x\rightarrow+\infty}\frac{\log f(x)}{\log g(x)}=c>a$,
there exists $x_{1}$ such that $\frac{\log f(x)}{\log g(x)}>a$ whenever $x>x_{1}$.
Since $g(x)\rightarrow+\infty$, there exists $x_{2}$ such that $\log g(x)>0$
whenever $x>x_{2}$. Let $x_{3}=\max(x_{1},x_{2})$. For any $x>x_{3}$,
we have that $\frac{\log f(x)}{\log g(x)}>a$ and $\log g(x)>0$.
Hence, $\log f(x)>a\log g(x)=\log\left(g(x)^{a}\right)$ . Since $\exp$
is strictly increasing, we further have $\exp\left(\log f(x)\right)>\exp\left(\log\left(g(x)^{a}\right)\right)$,
i.e., $f(x)>g(x)^{a}$. It follows that $\frac{f(x)}{g(x)}>\frac{g(x)^{a}}{g(x)}=g(x)^{a-1}\rightarrow+\infty$
as $x\rightarrow+\infty$. Therefore $\frac{f(x)}{g(x)}\rightarrow +\infty$ as $x\rightarrow +\infty$.
