Why is $\ln (\ln (n) - \ln (\ln (n)))$ asymptotically $\ln( \ln (n))$?

Question

So I sort of understand heuristically that $$\ln (\ln (n) - \ln (\ln (n))) = \ln(\ln(n)(1 - o(1))).$$ I can use laws of logs to get $$\ln(\ln(n)) + \ln(1 - o(1)).$$ This gets me what I want: $$\ln(\ln(n)) - o(1).$$ Which I believe shows that $\ln (\ln (n) - \ln (\ln (n)))$ is indeed asymptotically $\ln( \ln (n))$ (assuming that I am understanding little $o$ notation correctly).

What I am asking is if there is a more rigorous justification for the first equality stated. Does it suffice to simply show that $$\lim_{n\to \infty} \dfrac{\ln(\ln(n))}{\ln(n)} = 0,$$ or is there something else to do?

Also, while I am asking, I have a fairly dumb question, I can simply write $x - o(1)$ as $x + o(1)$ since it still carries the same idea of the asymptotic, correct?

Answer 1

\begin{align} \lim_{n\to\infty}\frac{\ln\big(\ln(n)-\ln(\ln(n)\big)}{\ln(\ln(n))} &= \lim_{n\to\infty}\frac{ \ln(\ln(n)) + \ln\left(1-\frac{\ln(\ln(n))}{\ln(n)}\right)}{\ln(\ln(n))} = 1 + \lim_{n\to\infty}\frac{\ln\left(1-\frac{\ln(\ln(n))}{\ln(n)}\right)}{\ln(\ln(n))}\\ &\overset{(*)}{=} 1 + \ln\left(\lim_{n\to\infty} 1-\frac{\ln(\ln(n))}{\ln(n)}\right)\cdot\lim_{n\to\infty}\frac{1}{\ln(\ln(n))}\\ &= 1+\ln(1)\cdot 0 = 1 \end{align} (*) follows by continuity. Therefore $\ln\big(\ln(n)-\ln(\ln(n))\big)\in \Theta\big(\ln(\ln(n))\big)$.

Answer 2

Because $$\frac{\ln(\ln{n}-\ln\ln{n})}{\ln\ln{n}}=\frac{\ln\ln{n}+\ln\left(1-\frac{\ln\ln{n}}{\ln{n}}\right)}{\ln\ln{n}}=$$ $$=1+\frac{\ln\left(1-\frac{\ln\ln{n}}{\ln{n}}\right)}{\ln\ln{n}}\rightarrow1$$ because $$\frac{\ln\ln{n}}{\ln{n}}=\ln\left(\ln{n}\right)^{\frac{1}{\ln{n}}}\rightarrow\ln1=0.$$