What is $\lim_{n\to \infty}\log(|\sin(\frac{1}{n})|)/\log(n+1)$

Question

Find:

$$\lim_{n \to \infty} \frac{\log(|\sin(\frac{1}{n})|)}{\log(n+1)}$$

I decided to use continuous extension theorem to solve this indeterminate form. Is this rigorous and correct?

Define $$f_n(x)= \lim_{n \to \infty} \frac{\log(|\sin(x+\frac{1}{n})|)}{\log(n+1)}$$ (Sequence of functions is not needed, but IMHO let’s one see more clearly the methodology)

Consider interval $(0,1]$

In this interval, $|\sin(x+\frac{1}{n})|>0 $ for all $n$, as well as when $n\to \infty$.

So the point-wise limit function $$f(x)= \lim_{n \to \infty} \frac{\log(|\sin(x+\frac{1}{n})|)}{\log(n+1)} = 0$$

As $f(x) = 0$ is constant on interval $(0,1]$, it is uniformly continuous. So using continuous extension theorem: [ Taken from page 10 of this book

We must have a unique extension given by, $\bar f(0) = \lim_{x \to 0} f(x) = 0$.

Therefore,

$$\lim_{n \to \infty} \frac{\log(|\sin(\frac{1}{n})|)}{\log(n+1)}=0$$

I would like to emphasize: there may be other better ways to solve this, but I’m interested in finding if this particular method is rigorous and correct. and if there is any issues with it, could it be modified slightly to fix such issues.

Thanks in advance!

Answer 1

For large $n$, we have $\sin(1/n)\approx 1/n$, so $$\log\sin 1/n\approx \log 1/n=-\log n$$ so the limit is $-1$