How to compute $\lim_{x\to0^-}{{\frac{\ln(\cot(-3x))}{\ln(\cot(-2x))}}}$? Is there any trick to evaluate this limit? I replace $\cot x = \sin x / \cos x$, but what next?
$$ \lim_{x\to0^-}{{\frac{\ln(\cot(-3x))}{\ln(\cot(-2x))}}}$$ 
 A: Hint: Write for $y>0$ close to $0$:
$$\log(\cot(y))=\log \cos(y)-\log(\frac{\sin(y)}{y})-\log(y)$$
use  that $\displaystyle \frac{\sin(y)}{y}\to 1$ as $y\to 0$ and apply for $y=-2x$, $y=-3x$. For $y=-2x$, this gives you that $\log(\cot(-2x))\sim -\log(-x)$, the same for $3x$. 
A: Let's put $x = -y$ so that $y \to 0^{+}$ and the function $f(x)$ changes to \begin{align}
g(y) &= \frac{\log\cot 3y}{\log \cot 2y}\notag\\
&= \frac{\log\cos 3y - \log\sin 3y}{\log\cos 2y - \log\sin 2y}\notag\\
&= \frac{\log\cos 3y - \log(\sin 3y/3y) - \log 3y}{\log\cos 2y - \log(\sin 2y/2y) - \log 2y}\notag\\
&= \frac{\log\cos 3y - \log(\sin 3y/3y) - \log 3 - \log y}{\log\cos 2y - \log(\sin 2y/2y) - \log 2 - \log y}\notag\\
&= \dfrac{\dfrac{\log\cos 3y}{\log y} - \dfrac{\log(\sin 3y/3y)}{\log y} - \dfrac{\log 3}{\log y} - 1}{\dfrac{\log\cos 2y}{\log y} - \dfrac{\log(\sin 2y/2y)}{\log y} - \dfrac{\log 3}{\log y} - 1}\notag\\
\end{align}
The answer is now obvious if we note that as $y \to 0^{+}$ we have $\log y \to -\infty$ and $$\cos 2y \to 1, \cos 3y \to 1, \sin 2y/2y \to 1, \sin 3y / 3y \to 1$$ and hence the first three terms in the numerator as well as in the denominator of $g(y)$ tend to $0$. The final answer is $-1/-1 = 1$.
A: You can use Hospital's rule to get
$$\eqalign{
  & \mathop {\lim }\limits_{x \to 0} {{\ln \left( {\cot \left( { - 3x} \right)} \right)} \over {\ln \left( {\cot \left( { - 2x} \right)} \right)}} = \mathop {\lim }\limits_{x \to 0} {{{{\left[ {\ln \left( {\cot \left( { - 3x} \right)} \right)} \right]}^\prime }} \over {{{\left[ {\ln \left( {\cot \left( { - 3x} \right)} \right)} \right]}^\prime }}} = \mathop {\lim }\limits_{x \to 0} {{{{{{\left[ {\cot \left( { - 3x} \right)} \right]}^\prime }} \over {\cot \left( { - 3x} \right)}}} \over {{{{{\left[ {\cot \left( { - 2x} \right)} \right]}^\prime }} \over {\cot \left( { - 2x} \right)}}}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{{{\left[ {\cot \left( { - 3x} \right)} \right]}^\prime }} \over {{{\left[ {\cot \left( { - 2x} \right)} \right]}^\prime }}}{{\cot \left( { - 2x} \right)} \over {\cot \left( { - 3x} \right)}} = \mathop {\lim }\limits_{x \to 0} {{{{ - 3} \over {{{\sin }^2}\left( { - 3x} \right)}}} \over {{{ - 2} \over {{{\sin }^2}\left( { - 2x} \right)}}}}{{\cot \left( { - 2x} \right)} \over {\cot \left( { - 3x} \right)}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {3 \over 2}\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( { - 2x} \right)} \over {{{\sin }^2}\left( { - 3x} \right)}}{{\cot \left( { - 2x} \right)} \over {\cot \left( { - 3x} \right)}} = {3 \over 2}\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( { - 2x} \right)\cot \left( { - 2x} \right)} \over {{{\sin }^2}\left( { - 3x} \right)\cot \left( { - 3x} \right)}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {3 \over 2}\mathop {\lim }\limits_{x \to 0} {{\sin \left( { - 2x} \right)\cos \left( { - 2x} \right)} \over {\sin \left( { - 3x} \right)\cos \left( { - 3x} \right)}} = {3 \over 2}\mathop {\lim }\limits_{x \to 0} {{\sin \left( { - 2x} \right)} \over {\sin \left( { - 3x} \right)}}\mathop {\lim }\limits_{x \to 0} {{\cos \left( { - 2x} \right)} \over {\cos \left( { - 3x} \right)}}  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {3 \over 2}\left( {{2 \over 3}} \right)\left( 1 \right) = 1 \cr} $$ 
