Find $\lim_{x\to 0} \frac {\ln( \sin 3x)}{\ln (\sin x)}$

Question

Find $$\lim_{x\to 0} \frac {\ln (\sin 3x)}{\ln (\sin x)}$$ without using L'Hospital

Now for this I have tried substituting $\sin 3x=3\sin x-4\sin^3x$ and then used some log properties to simply it further but it did not turn out to be good. I also tried to use the facts that $\lim_{x\to 0} \frac {\sin x}{x}=1$ and $\lim_{x\to 0} \dfrac {\ln (1+x)}{x}=1,$ but still had no success.

Any help would be greatly appreciated

Answer 1

Hint (no need of $\lim_{x\to 0} \frac {\sin x}{x}=1$ or $\lim_{x\to 0} \frac {\ln (1+x)}{x}=1$).

Note that $\sin(3x)= 3\sin(x) - 4\sin^3(x)$ (see HERE). Therefore, by letting $t=\sin(x)$, $$\lim_{x\to 0^+} \frac {\ln (\sin 3x)}{\ln (\sin x)}=\lim_{t\to 0^+} \frac {\ln (3t - 4t^3)}{\ln (t)}=\lim_{t\to 0^+} \frac {\ln(t)+\ln (3 - 4t^2)}{\ln (t)}=\lim_{t\to 0^+} \left( {1+\frac{\ln (3 - 4t^2)}{\ln(t)}}\right).$$

Answer 2

Note that for $x\to 0^+$ (otherwise the expression is not well defined)

$$\frac {\ln (\sin 3x)}{\ln (\sin x)}=\frac {\ln (\sin 3x)-\ln 3x+\ln3x}{\ln (\sin x)-\ln x+\ln x}=\frac {\ln (\frac{\sin 3x}{3x})+\ln3+\ln x}{\ln (\frac{\sin x}{x})+\ln x}\sim\frac{\ln x}{\ln x}\to 1$$

Answer 3

I think you can use Taylor expansion of $\sin x$: $$\sin (\alpha x)=\alpha x\ +\ o(x^2)$$ Then $$\lim_{x\to 0^+} \frac {\ln (\sin 3x)}{\ln (\sin x)}=\lim_{x\to 0^+} \frac {\ln 3x}{\ln x}=\lim_{x\to 0^+} \frac{\ln 3}{\ln x}+\frac{\ln x}{\ln x}=1$$