# How to prove which function is “stronger”?

I want to use $\log(x^3) < x^2$ so I compute:

$$\lim_{x\to+\infty}\frac{\log(x^3)}{x^2}$$

and with L'Hôpital showed it is equal to $0$; this implies that from certain point the function is "stronger". Is that enough?

• You can just show once and for all (with "L'Hopital" or something else) that $\frac{\ln x}{x} \xrightarrow[x\to\infty]{}0$. Then, you have $\frac{\ln x^3}{x^2} = \frac{3}{2}\cdot \frac{\ln x^2}{x^2}$, so you can reuse this result: the limit will follow. But if you want to know when the inequality will hold (this just will show it holds asymptotically, i.e. "for $x > C$ for some absolute constant $C$"), then you need a bit more work. – Clement C. Mar 16 '16 at 16:24
• You want to "use" log(x^3) < x^2 or you want to "show" it? – fleablood Mar 16 '16 at 16:42
• The way I understand it: OP wants to show $\log(x^3) < x^2$ is true, so that he can then use this inequality for ... (not specified). – StackTD Mar 16 '16 at 16:43
• Showing this limit is 0 would tell you that the inequality holds for sufficiently large values of $x$, not that it holds for all $x$. So the question is: what do you want to use / need to know, exactly? – StackTD Mar 16 '16 at 16:51

If you want to show that $\log(x^3) < x^2$ holds for all $x$, checking the behaviour for $x \to +\infty$ is not enough (*).
Consider the function ($x>0$): $$f(x)=x^2-\log(x^3) = x^2-3\log x$$ The derivative becomes $0$ at $x = \sqrt{3/2}$ and switches from negative to positive, so $f$ attains a minimum there. It is a global minimum so you have for all $x>0$: $$x^2-\log(x^3) \ge \underbrace{f(\sqrt{3/2})}_{\approx 0.89} > 0 \Rightarrow x^2 > \log(x^3)$$
(*) E.g. $\tfrac{x+2}{x^2} \to 0$ so $x^2 > x+2$ for sufficiently large $x$, but $x^2 \le x+2$ for $-1 \le x \le 2$.