Proof by induction: base case for a limit as x goes to infinity?

Question

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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@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 $x\to\infty$. — NoNames, Feb 09 '21 at 19:46
@Yves Daoust That would be an option, but I try to avoid l'Hospital, mostly because nobody ever checks whether it's applicable in the problem at hand (and it can give funny results if it isn't). — NoNames, Feb 09 '21 at 20:23
@NoNames: this is not what I mean. I mean that this is what the exercise is expecting. — , Feb 09 '21 at 20:48

score 0 · Answer 1 · answered Feb 09 '21 at 20:24

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.

score 0 · Answer 2 · answered Feb 09 '21 at 20:50

0

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.

answered Feb 09 '21 at 20:50

Proof by induction: base case for a limit as x goes to infinity?

2 Answers2