Does this number blow up or go to zero?

Question

$2\pi \epsilon^{3}(\log(\epsilon)+2\pi)^2$

Let $\epsilon \to 0$. I believe $2\pi \epsilon^{3}(\log(\epsilon)+2\pi)^2 \to 0$, and there is no concern about the blow-up of log at the origin. But can I be more specific and say that this number goes to zero like $\epsilon^3$? I.e., this number is in O($\epsilon^3)$.

What if the exponent on log is very large (but fixed)? Does it matter?

Thanks,

Answer 1

Given any fixed $t > 0$, for $\epsilon$ small enough you will have $|\ln(\epsilon)| < \epsilon^{-t}$, which ensures your limit is zero and that it goes to zero faster that $\epsilon^{3 - \delta}$ for any positive $\delta$ (but not $\delta = 0$).

One way to see the above inequality is to write $\epsilon = 1 / x$ and apply l'Hopital's rule on ${\displaystyle {\ln x \over x^t}}$ as $x \rightarrow \infty$.

Answer 2

Hint

Let us make it more general considering $$A=x^m\big(a+b \log(x)\big)^n$$ with $m,n>0$.

If you expand the term $\big(a+b \log(x)\big)^n$ using the binomial theorem, you will have a summation of terms $x^m \log^k(x)$ with $0\leq k\leq n$ and, by the knwon properties, each term will tend to $0$ when $x\to 0$.

So, for any pair $m,n>0$, $\lim_{x\to 0} A =0$