My analysis book [1] explains that $$ \lim_{x \to +\infty} \frac{(\log_a x)^α}{x^β} = 0 $$ for $β > 0$, and that by replacing $x = 1/y$ we get $$ \lim_{y \to 0^+}y^β(-\log_a y)^α = 0.\tag{*}\label{*} $$ Then it asks why it would be wrong to write $(\log_a y)$ instead of $(-\log_a y)$.
My reasoning was, well, I'd miss a factor $(-1)^α$, which changes the sign of the expression when $α$ is odd. Maybe I can divide both sides by $(-1)^α$ and, noting that as $y \to 0^+$
- $y^β$ is positive and
- $(\log_a y)^α$ is positive for even $α$, negative for odd $α$,
I'd get $$ \lim_{y \to 0^+}y^β(\log_a y)^α = \begin{cases} 0^+\cdot 1 = 0^+ & \text{for even $α$}\\ 0^-\cdot -1 = 0^+ & \text{for odd $α$} \end{cases} $$ and there I go, wrong result: the sign is flipped for odd $α$, and I don't know what it should be for even values of $α$. Where am I wrong?
Edit: I was wrong in taking for granted that $α$ is integer, which it is not, so my handling of $(-1)^α$ is wrong (thanks geetha290krm).
Still, is there an answer to the question the book poses in terms of effects of the minus sign on the expression? The fact that they asked it makes me wonder what else there is to be seen other than that $\log(1/y) = -\log y$.
1 Bramanti, Pagani, Salsa; Analisi 1 (2014), p. 132
In my opinion, it has nothing to do with odd/even. When you’re substituting $y$ in place of $x$ in the part $\log_ax$, what you are actually doing is $$\log_ax=\log_a\frac1y=\log_ay^{-1}=-\log_ay$$
And that’s where the minus sign comes from.