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

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

• Is it $\ln_{\gamma}(n)$ or $\ln^{\gamma}(n)$, your title and question differ... – Michael Burr Feb 14 '16 at 23:44

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 $\lim_{x\rightarrow\infty}e^{f(x)}$.
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.
• @ClementC. $(\log n)^\gamma \ne \gamma \log n$. Perhaps the OP meant $\log(n^\gamma)=\gamma \log(n)$. – Mark Viola Feb 15 '16 at 5:27