# $\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)}}=c$ 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 $$\displaystyle\lim_{x\to+\infty} \frac{f(x)}{g(x)}=+\infty$$

• If $c>1$, then there is some $\varepsilon > 0$ such that $c-\varepsilon = a > 1$. By the assumption, for $x$ big enough you have $\log f(x) \ge a \log g(x)$. What does this inequality tells you about $f$ and $g$ ? Commented Nov 9, 2022 at 19:01
• $f(x) = g(x) = 2$ is a counter-example. Commented Nov 9, 2022 at 19:36
• @John_Krampf: It is assumed that $\lim_{x\to+\infty} g(x)=+\infty$. Commented Nov 9, 2022 at 19:38
• @MartinR : Still, that condition is stated in the bottom paragraph, but not the top one. nn
– MSIS
Commented Dec 27, 2022 at 21:18
• @CPMP: Id suggest including the condition $lim_{x \rightarrow \ifty} g(x) = \infty$ in the top paragraph. Or at least that neither f,g are constant . Otherwise, f(x)-$e^c ; g(x)=e$ is a counterexample. It's not clear you're assuming that.
– MSIS
Commented Dec 27, 2022 at 21:22

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 stronger conclusion can be proved. For $$1\le a we get $$\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.$$

• Not really stronger (IMO) since $\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)}}=c$ is equivalent to $\lim_{x\to+\infty} \frac{\log{f(x)}}{\log{g(x)^a}}=c/a$. Commented Dec 27, 2022 at 21:34
• @MartinR That's right, neither imaginary stronger. Commented Dec 27, 2022 at 21:47

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 $$1. Since $$\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$$.