$$\mathop {\lim }\limits_{x \to \infty } \frac{{2\ln (x) + 3}}{{3{x^{\frac{1}{3}}}}}$$

How do you find the limit without using LHR? It's the classic $\infty \over \infty$.
I think the denominator is getting large "faster" than the numerator. Other than that, I'm not sure how to calculate a limit in this form. Maybe manipulating the $ln(x)$ in some way?


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    $\begingroup$ Write $t = \log x$. Ignoring the irrelevant constants, you have $$\lim_{t\to\infty} te^{-t/3}.$$ You know that $e^s$ grows faster than $s^2$. $\endgroup$ – Daniel Fischer Jan 5 '14 at 15:21
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    $\begingroup$ Your intuition is correct. For any $\alpha >0$, it holds that $\lim \limits_{x\to +\infty}\left(\ln(x)/x^\alpha\right)=0$ $\endgroup$ – Git Gud Jan 5 '14 at 15:22
  • $\begingroup$ @DanielFischer, would you like writing it as an answer? Thanks $\endgroup$ – AndrePoole Jan 5 '14 at 17:10

Writing $t = \log x$, we obtain

$$\lim_{t\to\infty} \frac{2t+3}{3e^{t/3}}.$$

The Taylor expansion of the exponential function immediately yields $e^s > \frac12 s^2$ for $s > 0$, and inserting that, we obtain the majorisation


that is easily seen to converge to $0$. On the other hand, $0$ is a lower bound for the expression, so

$$\lim_{t\to\infty} \frac{2t+3}{3e^{t/3}} = 0$$



You can use that $x^a$ grows faster than $\ln(x)$ for any $a>0$, and thus $$\lim_{x\to \infty}\frac{\ln(x)}{x^a}=0$$

  • $\begingroup$ What is radius of convergence for Taylor polynomial of $\ln(x)$? You cannot do Taylor expansion for $\ln(x)$ since it's radius of convergence is not $\infty$ so we cannot apply it to this limit. $\endgroup$ – neelp Jan 5 '14 at 15:26
  • $\begingroup$ @neelp I don't know what the radius of convergence is, but I removed that part. $\endgroup$ – Ragnar Jan 5 '14 at 15:31
  • $\begingroup$ @Ragnar The radius of convergence is the radius in which a series converges around the center. For $\ln$ and any center $x_0$, this radius is $<\infty$ and thus the series cannot be used for the limit $x\to\infty$ since it will diverge from the approximated function. $\endgroup$ – AlexR Jan 5 '14 at 15:35
  • $\begingroup$ @AlexR, I read the article on wikipedia and made a plot with Mathematica. It is pretty clear now that no taylor expansion of the $\ln$ can accurately describe the behaviour of it when $x\to \infty$. $\endgroup$ – Ragnar Jan 5 '14 at 15:37
  • $\begingroup$ @Ragnar: There is a Taylor expansion for $\ln(x+1) = 1-x+\frac{x^2}{2} - \frac{x^3}{3} + ...$. If you take enough terms, it will accurately describe the behaviour when $x\to \infty$. It is a very slow convergence, that's all. $\endgroup$ – user88595 Jan 5 '14 at 16:03

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