I haven't practiced limits for years, now I need them to solve an exercise and I don't know whether I have come up with the right solution.

$$\lim_{n \to \infty}{\frac{n^a}{\log\left(\left| \log(n^a)\right|\right)}}$$

where $a$ is a fixed constant.

Since I have the form $\frac{\infty}{\infty}$, I apply the De L'Hopital theorem, so I derive both numerator and denominator, so:

$$\lim_{n \to \infty}{\frac{an^{a-1}}{\frac{a}{n\log(n^a)}}} = \lim_{n \to \infty}{n\log(n^a)} = \lim_{n \to \infty}{a n\log(n)} = \infty$$

Can you please give me any feedback?

  • 1
    $\begingroup$ You're doing wrong with the derivative of the denominator. Rewrite it as $\log|a|+\log\log n$ (for $a\ne0$; but the expression is undefined for $a=0$). $\endgroup$
    – egreg
    Oct 7, 2013 at 15:24
  • $\begingroup$ So basically I am back to square one with $\frac{\infty}{\infty}$, shall I derive again? $\endgroup$
    – haunted85
    Oct 7, 2013 at 15:27
  • $\begingroup$ I meant: $\log(|\log(n^a)|)=\log|a|+\log\log n$; apply l'Hôpital to $\dfrac{n^a}{\log|a|+\log\log n}$, which is much easier. But it's not necessarily $\infty/\infty$, do you see why? $\endgroup$
    – egreg
    Oct 7, 2013 at 15:31
  • $\begingroup$ $\dfrac{n^a}{\log|a|+\log\log n} = \dfrac{an^a}{\log(n)}$, there must be something I am missing... $\endgroup$
    – haunted85
    Oct 7, 2013 at 15:37

1 Answer 1


The expression is defined only for $a\ne0$, because for $a=0$ the denominator would have $\log 0$.

So, assume $a>0$; then you can rewrite the denominator as $$ \log(|\log n^a|)=\log(|a\log n|)=\log(|a|\log n)=\log|a|+\log\log n $$ (at least for $n>1$, which is implied).

Since both numerator and denominator go to infinity, you can apply l'Hôpital's theorem: $$ \lim_{n\to\infty}\frac{n^a}{\log|a|+\log\log n}= \lim_{n\to\infty}\frac{an^{a-1}}{\frac{1}{\log n}\frac{1}{n}}= \lim_{n\to\infty}an^{a-1}\cdot n\log n= \lim_{n\to\infty}an^a\log n $$ provided this last limit exists. Does it?

Don't forget to apply the chain rule when differentiating $\log\log n$.

For $a<0$ you have a slightly different, but easier, situation.


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