Order of $h(x)=\frac{\ln^5(1+x\ln^{1\over8}(x))}{x^x-1}$

Question

I have to determine the order of infinitesimal $h(x)=\frac{\ln^5(1+x\ln^{1\over8}(x))}{x^x-1}$ with respect to $x$ for $x\to0^+$
$\ln^5(1+x\ln^{1\over8}(x))\sim (x\ln^{1\over8}(x))=x^5\ln^{5\over8}(x)$ and $x^x-1=e^{x\ln x}-1\sim x\ln x$.So we have : $$\lim_{x\to0^+}\frac{x^4}{\ln^{3\over8}(x)}$$I have been told that $4\lt ord(h(x))\lt 4+\epsilon$ ,where $\epsilon \in\mathbb{R}^+$.But how is it determined ?

Answer 1

Suppose $z\in\Bbb C$ and let $\varepsilon>0$. We aim to show that $$|z|^{4+\varepsilon}\ll |h(z)|\ll |z|^4\quad\text{as}\quad|z|\to0,$$where $$h(z)=\frac{\ln^5(1+z\ln^{1/8}(z))}{z^z-1}.$$ Using standard arguments (as you have done), we can show that $$h(z)\sim\frac{z^4}{\ln^{3/8}(z)}$$ for $|z|\to0$. Since $|\ln^{3/8}(z)|\to\infty$ as $|z|\to 0$, it is straightforward to see that $|h(z)|\leq|z|^4$. On the other hand, we have $$\left|\frac{h(z)}{z^{4+\varepsilon}}\right|\sim\left|\frac{1}{z^\varepsilon\ln^{3/8}(z)}\right|.$$ Again, applying standard arguments, we can show the right-hand side tends to $\infty$ with $|z|\to0$. Since $\varepsilon>0$ was arbitrary, we have that the order of $|h(z)|$ is strictly greater than $|z|^{4+\varepsilon}$ for any fixed $\varepsilon>0$.