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I've been trying to compute the following limit for a few hours. Let $f(\gamma, \beta)$ be defined as follows:

$$f(\gamma, \beta)=\lim_{x \rightarrow \infty} (1-\gamma^{1/x})(\log(x))^{\beta}.$$

I am searching for a function $g$ so that for all $\gamma$,

$$0<f(\gamma, g(\gamma))<\infty.$$

I have done some numerical computations and checked that $g \neq \operatorname{id}$, and tried L'Hôpital's rule and various comparisons. If anyone has seen something like this before I would appreciate hearing.

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    $\begingroup$ This is really calculus, shouldn't be in MO. $\endgroup$
    – username
    Commented May 30, 2022 at 10:52
  • $\begingroup$ OK! Can you recommend a more suitable venue? $\endgroup$
    – colin
    Commented May 30, 2022 at 18:28
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    $\begingroup$ The Mathematics Stack Exchange $\endgroup$
    – username
    Commented May 30, 2022 at 18:36
  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Jun 24, 2022 at 15:12

1 Answer 1

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We have $f(\gamma,\beta)=0$ for every $\gamma>0$ and $\beta\in\mathbb{R}$. Indeed, $1-\gamma^{1/x}$ is asymptotically $(\log\gamma)/x$, and $(\log x)^\beta/x$ tends to zero for any $\beta\in\mathbb{R}$ (as $x$ tends to infinity).

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