Find $\lim_{x\to 0} \frac {\ln( \sin 3x)}{\ln (\sin x)}$ 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
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 $t=\sin(x)$,
$$\lim_{x\to 0^+} \frac {\ln (\sin 3x)}{\ln (\sin x)}=\lim_{t\to 0^+} \frac {\ln (3t - 4t^3)}{\ln (t)}=\lim_{t\to 0^+} \frac {\ln(t)+\ln (3 - 4t^2)}{\ln (t)}=\lim_{t\to 0^+} \left( {1+\frac{\ln (3 - 4t^2)}{\ln(t)}}\right).$$
I think you can use Taylor expansion of $\sin x$:
$$\sin (\alpha x)=\alpha x\ +\ o(x^2)$$
Then
$$\lim_{x\to 0^+} \frac {\ln (\sin 3x)}{\ln (\sin x)}=\lim_{x\to 0^+} \frac {\ln 3x}{\ln x}=\lim_{x\to 0^+}  \frac{\ln 3}{\ln x}+\frac{\ln x}{\ln x}=1$$