Evaluation of $\lim_{x\rightarrow 0}\frac{\ln(\sin 2x)}{\ln(\sin x)}$

Question

Evaluation of $$\lim_{x\rightarrow 0^{+}}\frac{\ln(\sin 2x)}{\ln(\sin x)}$$

without using without l'Hôpital's rule

$\bf{My\; Try::}$ Using $$\frac{2x}{\pi}x<\sin x<x\;\forall x \in \left(0,\frac{\pi}{2}\right)$$

So $$\ln\left(\frac{4x}{\pi}\right)<\ln(\sin 2x)<\ln(2x)$$

So $$\frac{\ln (4x)-\ln(\pi)}{\ln(\sin x)}<\frac{\ln (\sin 2x)}{\ln(\sin x)}<\frac{\ln(2x)}{\ln(\sin x)}$$

Now how can I solve it? Help required, Thanks

Answer 1

$$\lim_{x\to 0}\frac{\ln(\sin 2x)}{\ln(\sin x)}=\lim_{x\to 0}\frac{\ln(2\sin x\cos x)}{\ln(\sin x)}=\lim_{x\to 0}\frac{\ln(\sin x)+\ln (2\cos x)}{\ln(\sin x)}$$ $$=\lim_{x\to 0}\left(1+\frac{\ln (2\cos x)}{\ln(\sin x)}\right)=1+0=1$$

Answer 2

From the duplication formula,

$$\ln\sin2x=\ln2+\ln\sin x+\ln\cos x$$

then after discarding the bounded terms, your limit to $0^+$ is $1$.