Asymptotics of logarithm: $\frac{1}{n}\ln(a+o(1)) = \frac{1}{n}\ln(a)+o(\frac{1}{n})$

Question

I am having problems with the use of the little oh notation my professor is adopting in the solutions to some exercises.

As an example I do not understand why $$ \frac{1}{n}\ln(a+o(1)) = \frac{1}{n}\ln(a)+o(\frac{1}{n}). $$ It seems to me that the definition of little oh is not verified because $$ \lim_{n \to \infty} \frac{\frac{1}{n}\ln(a+o(1)) - \frac{1}{n}\ln(a))}{\frac{1}{n}} $$ does not equal $0$, I remain with $$ \lim_{n \to \infty} \ln(a+o(1)) - \ln(a)) $$ that I do not know exactly how to treat but it seems to me it can't equal $0$ because $o(1)$ is an unknown function that does not depend on $n$.

Answer 1

The notation $o(h)$ is not a specific function. In a limit settings, it is usually used to denote a small quantity compared to the quantity $h$.

In this given settings, one is working with $\lim\limits_{n\to\infty}$. Then one should take $o(1)$ as some quantity that $$\lim_{n\to\infty}\frac{o(\color{red}{1})}{\color{red}{1}}=0,$$ and $o(\frac1n)$ some (other) quantity satisfying $$\lim_{n\to\infty}\frac{o(\color{blue}{\frac1n})}{\color{blue}{\frac1n}}=0.$$ Inparticular, $$\frac{1}{n}o(1)=o(\frac1n).$$

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Back to your examples, the facts that $\ln$ is a continuous function and that $\lim\limits_{n\to\infty}o(1)=0$ imply $$\lim_{n\to\infty}\ln(a+o(1))=\ln a,$$ which is equivalent to the statement

If $x_n\to 0$ then $\lim\limits_{n\to\infty}\ln(a+x_n)=\ln a.$

and is equivalent to

$$\ln(a+o(1))-\ln a=o(1).$$

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Now can you work out the meaning of each statement in the proof?