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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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  • $\begingroup$ Welcome to MSE. Please type your questions instead of posting images. Images can't be browsed and are not accessible to those using screen readers. If you need help formatting math on this site, here's a tutorial $\endgroup$
    – saulspatz
    Feb 9, 2021 at 19:25
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    $\begingroup$ 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

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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.

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Hint:

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

$$\lim_{x\to\infty}\frac{\log^k(x)}{x}=\lim_{x\to\infty}\frac{k\log^{k-1}(x)}{x}.$$

Now the induction should be obvious.

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