limit $r^{n-1}(\log(1/r))^n$ as $r$ goes to zero for $n>1$

Question

I am trying to find the limit $r^{n-1}(\log(1/r))^{n}$ as $r$ goes to zero and $r\geq 0$

Attempt

Del' Hopitals for $\dfrac{r^{n-1}}{(\log(1/r))^{-n}}$ simply rehashes the same fraction up to a constant.

The classic logarithm inequality gives $r^{n-1}(\log(1/r))^{n}\leq r^{n-1}(\frac{1}{r}-1)^{n}=\frac{1}{r}(1-r)^{n}\to \infty$.

any suggestions?

Thanks

Answer 1

There is certainly an inequality in the other direction that you can use, but students are easily confused by logarithm inequalities. If this is the case for you, you may want to transform the limit to get rid of the logarithm and bring the expression to a more vivid form. Since $r$ tends to zero, we can write $r=e^{-s}$ where, in order for $r$ to go to zero, $s$ must tend to plus infinity. Then $$\lim_{r\to 0^+} r^{n-1}(\log(1/r))^n$$ will equal $$\lim_{s\to \infty}e^{-s(n-1)}s^n = \lim_{s\to \infty} \frac{s^n}{e^{s(n-1)}}.$$ Now what do you know about the behavior at infinity of polynomials versus exponentials? Can you finish the problem now?

Answer 2

$r^{n-1}\log^n(1/r)=(-1)^n r^{n-1}\log^n r\to0$ since $n>1$.