$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^\alpha} ; \quad \alpha >0$$

Wouldn't it be just zero because $x^\alpha$ will reach $\infty$ much earlier than $\log(x)$?

I tried using L'Hospital's rule but was not able to figure it. Any help will be appreciated.


  • 1
    $\begingroup$ L'Hopital should work just fine here $\endgroup$
    – bsbb4
    Jan 11, 2020 at 23:15
  • $\begingroup$ What did you get when you applied L'Hopital's Rule? It sounds as if you made an error in application or interpreting the result, but there is no way to help with that because you have chosen to hide your work and result. $\endgroup$ Jan 11, 2020 at 23:50
  • $\begingroup$ I got the same result as David. $\endgroup$ Jan 11, 2020 at 23:51
  • $\begingroup$ Jut to clarify, the equation $\lim\limits _{x\to \infty} \dfrac{\log x} {x^a} =0,a>0$ is expressed in informal/crude language as "logarithm goes to infinity much slower than any polynomial function" and hence the latter can't be used as a justification for the former. $\endgroup$
    – Paramanand Singh
    Jan 12, 2020 at 8:05

2 Answers 2


Assuming $\log$ means the natural logarithm, L'Hospital's gives

$$\lim_{x \rightarrow \infty} \frac{\log(x)}{x^\alpha }=\lim_{x \rightarrow \infty} \frac{1/x}{\alpha x^{\alpha-1} }=\dfrac{1}{\alpha}\lim_{x \rightarrow \infty} \frac{1}{x^{\alpha} }$$

  • $\begingroup$ I also got the same thing but what can I do now? $\endgroup$ Jan 11, 2020 at 23:10
  • $\begingroup$ @Leaderboard281923 : So the limit is $0. \qquad$ $\endgroup$ Jan 11, 2020 at 23:11

Without L'Hospital:

Any high-school student has learnt that $\;\lim_{x\to\infty}\dfrac{\ln x}x=0$ (which is proved without the help of the not-so-divine Marquis…).

Now $\ln x=\frac 1\alpha\ln(x^\alpha)$, so, setting $u=x^\alpha$, $\lim_{x\to \infty}u=\infty$, and $$\frac{\ln x}{x^\alpha}=\frac1\alpha\frac{\ln u}u\xrightarrow[u\to \infty]{}\frac1\alpha\cdot 0=0.$$

In the same vein, it is easy to show that for any $\alpha,\beta>0$, one has

$\qquad\qquad\displaystyle\lim_{x\to\infty}\frac{\ln^\beta > x}{x^\alpha}=0,\quad\text{i.e.}\quad \ln^\beta > x=_{\infty}o\bigl(x^\alpha\bigr).$

  • $\begingroup$ In fact, this is true for all $\beta\in\mathbb{R}, \alpha>0$. $\endgroup$
    – bjorn93
    Jan 12, 2020 at 1:18
  • $\begingroup$ @bjorn93: Yes, but if $\beta\le 0$, we don't have to state it as a result – it is no more an indeterminate form. $\endgroup$
    – Bernard
    Jan 12, 2020 at 10:26

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