With induction we always start with a base case; What would the base case for this be? Choosing 1 seems nonsensical. Choosing infinity seems wrong as well.

Prove, using induction, that $\lim\limits_{x\to\infty}\dfrac{(\ln x)^k}x=0$.

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    – saulspatz
    Feb 9, 2021 at 19:25
  • 3
    $\begingroup$ The induction is on $k$, and the base case is $k=0$. $\endgroup$ Feb 9, 2021 at 19:25
  • $\begingroup$ @Brian M. Scott That's a reasonable guess, assuming that $k$ is a variable taking non-negative integer values (the OP doesn't tell us, unfortunately). On the other hand, induction doesn't seem very natural, since the induction step multiplies the expression by $\ln x$, and that tends to $\infty$ as $x\to\infty$. $\endgroup$
    – NoNames
    Feb 9, 2021 at 19:46
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    $\begingroup$ @saulspatz Yeah.... first post... dummy mistake $\endgroup$ Feb 9, 2021 at 19:52
  • $\begingroup$ @NoNames: my bet is that L'Hospital needs to be used. $\endgroup$
    – user65203
    Feb 9, 2021 at 19:53

2 Answers 2


As I said in a comment, it's a strange idea, but let's give it a try: so we want to prove $$\lim_{x\to\infty} \frac{\ln^k x}x=0$$ for $k=0,1,\ldots$ This is obvious for $k=0$. Defining $$f_k(x)=\frac{\ln^k x}x,$$ it isn't a good idea to use $f_{k+1}(x)=\ln x\,f_k(x)$ for the induction step, but $$f_{k+1}(x)=2^{k+1}\,f_k(\sqrt{x})\frac{\ln\sqrt{x}}{\sqrt{x}}<2^{k+1}\,f_k(\sqrt{x})$$ will turn the trick.



By L'Hospital, for $k>0$,


Now the induction should be obvious.


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