Limit Involving Composition of $\ln$ and $\sin$ I have problem that find:
$$\lim_{x\rightarrow 0}\frac{\ln(\sin(mx))}{\ln(\sin(nx))}.$$
I do not know how to find it. Please show me a hint, or a solution is very appriciate. Thanks.
 A: Actually, it should be $\lim_{x \to 0+}$, not $\lim_{x \to 0}$, unless you want to deal with logarithms of negative numbers.
$\lim_{x \to 0} \sin(x)/x = 1$.  So write 
$$\ln(\sin(mx)) = \ln \left( m x \frac{\sin(mx)}{mx} \right) = \ln(x) + \ln(m) + \ln\left( \frac{\sin(mx)}{mx} \right)$$
and similarly for $\ln(\sin(nx))$.  As $x \to 0+$, $\ln(x) \to -\infty$ while $\ln(m)$ is constant and $\ln\left(\sin(mx)/(mx)\right)$ or $\ln\left(\sin(mx)/(mx)\right)$ goes to $\ln(1) = 0$. 
Divide both top and bottom by the main term $\ln(x)$:
$$ \dfrac{1 + \ln(m)/\ln(x) + \ln(\sin(mx)/(mx))/\ln(x)}{1 +\ln(n)/\ln(x) + \ln(\sin(nx)/(nx))/\ln(x)} \to \dfrac{1+0+0}{1+0+0} = 1$$ 
A: This limit (corrected to the limit as $x\to 0^+$) is also very straightforward to calculate using l’Hospital’s rule:
$$\begin{align*}
\lim_{x\to 0^+}\frac{\ln\sin mx}{\ln\sin nx}&=\lim_{x\to 0^+}\frac{\frac{m\cos mx}{\sin mx}}{\frac{n\cos nx}{\sin nx}}\\\\
&=\lim_{x\to 0^+}\frac{m\cot mx}{n\cot nx}\\\\
&=\lim_{x\to 0^+}\frac{m\tan nx}{n\tan mx}\\\\
&=\lim_{x\to 0^+}\left(\frac{m}{\tan mx}\cdot\frac{\tan nx}n\right)\\\\
&=1\cdot 1\\\\
&=1\;,
\end{align*}$$
since $\displaystyle\lim_{y\to 0}\frac{\tan y}y=1$.
