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I have the following problem to prove.

Let $\gamma, δ \in \mathbb{R}^\ast_+$ and $f\colon \mathbb{R}^\ast_+\to \mathbb{R}$ defined by $f(x) = \delta x − \gamma\ln x$.

Accepting that $\lim_{x\to+\infty }f(x) = +\infty$ prove that $\lim_{n\to\infty} \frac{n^\delta}{\ln^\gamma n} = +\infty$.

The following substitution is true: $$n^\delta=e^{\delta\ln n}$$

Any hint, suggestion or solution is welcomed.

Thank you

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    $\begingroup$ Is it $\ln_{\gamma}(n)$ or $\ln^{\gamma}(n)$, your title and question differ... $\endgroup$ – Michael Burr Feb 14 '16 at 23:44
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Hint: Consider $$ e^{f(x)}=e^{\delta x-\gamma\ln x}. $$ Choose $x=\ln n$, then $$ e^{f(x)}=e^{\delta x-\gamma\ln x}=e^{\delta\ln n-\gamma\ln \ln n}=\frac{e^{\delta\ln n}}{e^{\gamma\ln \ln n}}=\frac{n^\delta}{(\ln n)^\gamma}. $$ Since $n\rightarrow \infty$, $\ln n\rightarrow\infty$ and so you can relate what you have to $\lim_{x\rightarrow\infty}e^{f(x)}$.

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So you must show that $g(n)\stackrel{\rm def}{=}\frac{n^\delta}{(\ln n)^\gamma} \xrightarrow[n\to\infty]{} \infty$, with $\delta,\gamma > 0$.

This is equivalent to showing that $\ln g(n) \xrightarrow[n\to\infty]{} \infty$ (can you see why?).

But $\ln g(n) = \delta \ln n - \gamma\ln \ln n = f(\ln n)$. Use your hypothesis now.

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    $\begingroup$ @ClementC. $(\log n)^\gamma \ne \gamma \log n$. Perhaps the OP meant $\log(n^\gamma)=\gamma \log(n)$. $\endgroup$ – Mark Viola Feb 15 '16 at 5:27
  • $\begingroup$ Yes, I was missing a log here, in both places. Silly mistake... Thanks! (Fixed) $\endgroup$ – Clement C. Feb 15 '16 at 12:13

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