Find $\lim_{x\to 0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}$ without using the L'Hopital's rule or Taylor's series This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)},
$$
where $a,b=const$. I know how to calculate this limit using the L'Hopital's rule:
$$
\lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}=
\lim_{x\to0}\frac{\frac{2a\sin (ax)\cos (ax)}{\sin^2(ax)}}{\frac{2b\sin (bx)\cos (bx)}{\sin^2(bx)}}=
\lim_{x\to0}\frac{a}{b}\cdot\frac{\sin (ax)\cos (ax)}{\sin (bx)\cos (bx)}\cdot\frac{\sin^2(bx)}{\sin^2(ax)}
$$
(using the asymptotic equivalence $\sin x\sim x$)
$$
=\lim_{x\to0}\frac{a}{b}\cdot\frac{ax}{bx}\cdot\frac{(bx)^2}{(ax)^2}=1,
$$
but I don't know to calculate this limit without derivatives.
 A: Since, $\sin^2(x)=\sin^2(-x)$, we can accept $a,b\in \mathbb R^{+}$. Thus we have,
$$\begin{align}\lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}&=\lim_{x\to0}\frac{2\ln|\sin(ax)|}{2\ln|\sin(bx)|}\\
&=\lim_{x\to0}\frac{\ln|\sin(ax)|}{\ln|\sin(bx)|}\\
&=\lim_{x\to0^{+}}\frac{\ln\sin(ax)}{\ln\sin(bx)}\\
&=\lim_{x\to0^{+}}\left(\frac{\frac{\ln\sin(ax)}{\ln ax}}{\frac{\ln\sin(bx)}{\ln bx}}\times \frac{\ln ax}{\ln bx}\right)\\
&=1.\end{align}$$

Explanation:
Let $\alpha,\thinspace x\in\mathbb R^{+}$, then we can write the following limits:

*

*$\lim_{x\to0^{+}}\frac{\ln\sin(\alpha x)}{\ln\alpha x}=1$.

Use:
$\ln(\sin \alpha x) = \ln \alpha x + \ln\left(\frac {\sin \alpha x} {\alpha x}\right)$


*$\lim_{x\to 0^{+}}\frac{\ln ax}{\ln bx}=1$
Use:
$\ln (\alpha x)=\ln \alpha+\ln x.$
A: Can someone tell me what is wrong in what I'm doing?
Assuming $ \lim_{x\to0}\frac{\sin\left(\alpha x\right)}{\alpha x}=1 $
$$ \require{cancel} \lim_{x\to0}\frac{\ln\left(\sin^{2}\left(ax\right)\right)}{\ln\left(\sin^{2}\left(bx\right)\right)}=\frac{\ln\left(\frac{\sin^{2}\left(ax\right)}{\left(ax\right)^{2}}\cdot\left(ax^{2}\right)\right)}{\ln\left(\frac{\sin^{2}\left(bx\right)}{\left(bx\right)^{2}}\cdot\left(bx^{2}\right)\right)}=\frac{\ln\left(\cancelto{1}{\frac{\sin^{2}\left(ax\right)}{\left(ax\right)^{2}}}\right)+\ln\left(\left(ax\right)^{2}\right)}{\ln\left(\cancelto{1}{\frac{\sin^{2}\left(bx\right)}{\left(bx\right)^{2}}}\right)+\ln\left(\left(bx\right)^{2}\right)}=\frac{\ln\left(1\right)+\ln\left(\left(ax\right)^{2}\right)}{\ln\left(1\right)+\ln\left(\left(bx\right)^{2}\right)}=\frac{\ln\left(\left(ax\right)^{2}\right)}{\ln\left(\left(bx\right)^{2}\right)}\Rightarrow \\
\frac{2\ln\left(ax\right)}{2\ln\left(bx\right)}=\frac{\ln\left(ax\right)}{\ln\left(bx\right)}=\frac{\ln\left(a\right)+\ln\left(x\right)}{\ln\left(b\right)+\ln\left(x\right)} $$
Dividing both sides by $\ln(x) $ will bring us to $ \lim_{x\to0}\frac{\ln\left(a\right)+\ln\left(x\right)}{\ln\left(b\right)+\ln\left(x\right)}=\lim_{x\to0}\frac{\frac{\ln\left(a\right)}{\ln\left(x\right)}+\frac{\ln\left(x\right)}{\ln\left(x\right)}}{\frac{\ln\left(b\right)}{\ln\left(x\right)}+\frac{\ln\left(x\right)}{\ln\left(x\right)}}=\frac{0+1}{0+1}=1 $
