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


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

  • $\begingroup$ 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. $\endgroup$ – Clement C. Mar 16 '16 at 16:24
  • $\begingroup$ You want to "use" log(x^3) < x^2 or you want to "show" it? $\endgroup$ – fleablood Mar 16 '16 at 16:42
  • $\begingroup$ 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). $\endgroup$ – StackTD Mar 16 '16 at 16:43
  • $\begingroup$ 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? $\endgroup$ – StackTD Mar 16 '16 at 16:51

Do you only need the asymptotic behaviour? See comments.

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 $-1 \le x \le 2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.