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

Question

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.

Answer 1

Since, $\sin^2(x)=\sin^2(-x)$, we can accept $a,b\in \mathbb R^{+}$. Thus we have,

$$\begin{align}\lim_{x\to0}\frac{\ln\sin^2(ax)}{\ln\sin^2(bx)}&=\lim_{x\to0}\frac{2\ln|\sin(ax)|}{2\ln|\sin(bx)|}\\ &=\lim_{x\to0}\frac{\ln|\sin(ax)|}{\ln|\sin(bx)|}\\ &=\lim_{x\to0^{+}}\frac{\ln\sin(ax)}{\ln\sin(bx)}\\ &=\lim_{x\to0^{+}}\left(\frac{\frac{\ln\sin(ax)}{\ln ax}}{\frac{\ln\sin(bx)}{\ln bx}}\times \frac{\ln ax}{\ln bx}\right)\\ &=1.\end{align}$$

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Explanation:

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

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

Use:

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

2. $\lim_{x\to 0^{+}}\frac{\ln ax}{\ln bx}=1$

Use:

$\ln (\alpha x)=\ln \alpha+\ln x.$